MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 6

a, b

$a, b$ and

c

$c$ arc in arithmetic progression then

\frac{b - a}{c - b}

$\dfrac{b- a}{c - b}$ is equal to

0%
$\dfrac{b}{a}$
0%
$0$
0%
$1$
0%
$2a$

Explanation

Solution:

a, b

$a,b$ and

c

$c$ are in A.P. then,

b - a = c - b =

$b-a=c-b=$ common difference

or,

\frac{b - a}{c - b} = 1

$\cfrac{b-a}{c-b}=1$

Therefore,

C

$C$ is the correct option.

If the sum of an AP is the same for

p

$p$ terms as for the

q

$q$ terms , Find the sum for

(p + q)

$(p+q)$ terms

0%
$2$
0%
$0$
0%
$4$
0%
None of these

Explanation

Since

S_{p} = S_{q}

$S_p = S_q$

\frac{p}{2} [2 a + (p - 1) d] = \frac{q}{2} [2 a + (q - 1) d] 2 a p + (p - 1) p d = 2 a q + q (q - 1) d 2 a (p - q) = [q (q - 1) - p (p - 1)] d 2 a (p - q) = [q^{2} - p^{2} + (p - q)] d 2 a (p - q) = [(q - p) (q + p) + (p - q)] d 2 a = [- (q + p) + 1] d - 2 a = (q + p - 1) d ∴ S_{p + q} = \frac{(p + q)}{2} [2 a + (p + q - 1) d] S_{p + q} = \frac{(p + q)}{2} (2 a - 2 a) = 0

$\cfrac { p }{ 2 } [2a+(p-1)d]=\cfrac { q }{ 2 } [2a+(q-1)d]\\ 2ap+(p-1)pd=2aq+q(q-1)d\\ 2a(p-q)=[q(q-1)-p(p-1)]d\\ 2a(p-q)=[{ q }^{ 2 }-{ p }^{ 2 }+(p-q)]d\\ 2a(p-q)=[(q-p)(q+p)+(p-q)]d\\ 2a=[-(q+p)+1]d\\ -2a=(q+p-1)d\\ \therefore { S }_{ p+q }=\cfrac { (p+q) }{ 2 } [2a+(p+q-1)d]\\ { S }_{ p+q }=\cfrac { (p+q) }{ 2 } (2a-2a)=0$

Firs term of an arithmetic progression is

2

$2$ . If the sum of its first five terms is equal to one-fourth of the sum of the next five terms, then the sum of its first

30

$30$ terms is

0%
$2670$
0%
$2610$
0%
$-2520$
0%
$2550$

Explanation

Given,

First term

a = 2

$a=2$

According to the problem, first 10 are

a, a + d, a + 2 d, a + 3 d, a + 4 d, a + 5 d, a + 6 d, a + 7 d, a + 8 d, a + 9 d

$a,a+d, a+2d, a+3d,a+4d, a+5d, a+6d, a+7d, a+8d, a+9d$

5 a + 10 d = \frac{1}{4} (5 a + 35 d)

$5a+10d=\dfrac{1}{4}(5a+35d)$

20 a + 40 d = 5 a + 35 d

$20a+40d = 5a+35d$

15 a = - 5 d

$15a = - 5d$

3 a = - d

$3a = -d$

a = 2

$a=2$

d = - 6

$d =-6$ and

n = 30

$n=30$

S_{30} = \frac{n}{2} [2 (a) + (n - 1) d]

$S_{30} = \dfrac{n}{2}[2(a)+(n-1)d]$

S_{30} = \frac{30}{2} [2 (2) + (29) (- 6)]

$S_{30} = \dfrac{30}{2}[2(2)+(29)(-6)]$

= - 2550

$=-2550$

(n - 2)^{t h}

$(n-2)^{th}$ term of an arithmetic progression will be __________.

0%
$a+(n-1)d$
0%
$a+(n-3)d$
0%
$a+(n-2)d$
0%
None

Explanation

n^{t h}

$n^{th}$ term of an A.P. is

T_{n} = a + (n - 1) d

$T_n=a+(n-1)d$

∴ (n - 2)^{t h}

$\therefore (n-2)^{th}$ term of an A.P. is

T_{n - 2} = a + (n - 2 - 1) d

$T_{n-2}=a+(n-2-1)d$

= a + (n - 3) d

$=a+(n-3)d$

S_{r}

$S_{r}$ denotes the sum of

r

$r$ terms of an AP and

\frac{S_{a}}{a^{2}} = \frac{S_{b}}{b^{2}} = c

$\dfrac {S_{a}}{a^{2}} = \dfrac {S_{b}}{b^{2}} = c$ then

S_{c}

$S_{c}$ is

0%
$c^{3}$
0%
$c/ ab$
0%
$abc$
0%
$a + b + c$

Explanation

S_{r}

${ S }_{ r }$ denotes sum of

r

$r$ terms.

\frac{S_{a}}{a^{2}} = \frac{S_{b}}{b^{2}} \frac{\frac{a}{2} (2 a_{1} + (a - 1) d)}{a^{2}} = \frac{\frac{b}{2} (2 a_{1} + (b - 1) d)}{b^{2}} \frac{2 a_{1}}{a} + \frac{(a - 1) d}{a} = \frac{2 a_{1}}{b} + \frac{(b - 1) d}{b} o r, \frac{2 a_{1}}{a} + d - \frac{d}{a} = \frac{2 a_{1}}{b} + d - \frac{d}{b} o r, 2 a_{1} (\frac{1}{a} - \frac{1}{b}) = d (\frac{1}{a} - \frac{1}{b}) S o, \frac{a}{2} (\frac{2 a_{1} + (a - 1) d}{a^{2}}) = c o r, 2 a_{1} \frac{(1 + a - a)}{2 a} = c \frac{a_{1} a}{a} = c c = a_{1} S_{c} = \frac{c}{2} (2 a_{1} + (c - 1) d) = \frac{c}{2} (2 a_{1} + (c - 1) 2 a_{1}) = \frac{c}{2} (2 c + (c - 1) 2 c) = \frac{c}{2} \times 2 c (1 + c - 1) = c^{3}

$\cfrac { { S }_{ a } }{ { a }^{ 2 } } =\cfrac { { S }_{ b } }{ { b }^{ 2 } } \\ \cfrac { \cfrac { a }{ 2 } (2{ a }_{ 1 }+(a-1)d) }{ { a }^{ 2 } } =\cfrac { \cfrac { b }{ 2 } (2{ a }_{ 1 }+(b-1)d) }{ { b }^{ 2 } } \\ \cfrac { 2{ a }_{ 1 } }{ a } +\cfrac { (a-1)d }{ a } =\cfrac { 2{ a }_{ 1 } }{ b } +\cfrac { (b-1)d }{ b } \\ or,\cfrac { 2{ a }_{ 1 } }{ a } +d-\cfrac { d }{ a } =\cfrac { 2{ a }_{ 1 } }{ b } +d-\cfrac { d }{ b } \\ or,2{ a }_{ 1 }(\cfrac { 1 }{ a } -\cfrac { 1 }{ b } )=d(\cfrac { 1 }{ a } -\cfrac { 1 }{ b } )\\ So,\cfrac { a }{ 2 } (\cfrac { 2{ a }_{ 1 }+(a-1)d }{ { a }^{ 2 } } )=c\\ or,2{ a }_{ 1 }\cfrac { (1+a-a) }{ 2a } =c\\ \cfrac { { a }_{ 1 }a }{ a } =c\\ c={ a }_{ 1 }\\ { S }_{ c }=\cfrac { c }{ 2 } (2{ a }_{ 1 }+(c-1)d)\\ =\cfrac { c }{ 2 } (2{ a }_{ 1 }+(c-1)2{ a }_{ 1 })\\ =\cfrac { c }{ 2 } (2c+(c-1)2c)\\ =\cfrac { c }{ 2 } \times 2c(1+c-1)\\ ={ c }^{ 3 }$

Answer

(A)

$(A)$

Find the sum of the first

25

$25$ terms of an A.P whose

n

$n$ th term is given by

a_{n} = 8 - 3 n

${a}_{n}=8-3n$

0%
$-775$
0%
$-500$
0%
$800$
0%
$500$

Explanation

a_{n} = 8 - 3 n

$a_{n}=8-3n$

a_{1} = 5

$a_{1}=5$

a_{2} = 2

$a_{2}=2$

d = 2 - 5 = - 3

$d=2-5=-3$

Thus,

S_{25} = \frac{25}{2} [10 + 24 (- 3)]

$S_{25}=\dfrac{25}{2}[10+24(-3)]$

= \frac{25}{2} \times - 62 = - 775

$=\dfrac{25}{2} \times -62=-775$

Hence, option A is correct.

There are

20

$20$ rows of seats in a conference hall with

20

$20$ seats in the first row,

21

$21$ seats in the second row,

22

$22$ seats in the third row and so on. In total, how many seats are there in the conference hall?

0%
$500$
0%
$550$
0%
$590$
0%
$620$

Explanation

Given:-

Total number of rows

(n) = 20

$(n) = 20$

Seats in first row

(a) = 20

$(a) = 20$

Seats in second row

(a_{2}) = 21

$({a}_{2}) = 21$

Seats in third row

(a_{3}) = 22

$({a}_{3})= 22$

Common difference

(d) = a_{3} - a_{2} = 22 - 21 = 1

$(d) = {a}_{3} - {a}_{2}= 22-21 = 1$

\Rightarrow Total number of seats in conference hall (S_{n}) = \frac{n}{2} (2 a + (n - 1) d)

$\Rightarrow \text{Total number of seats in conference hall } ({S}_{n}) = \cfrac{n}{2} \left(2a + \left(n-1\right)d \right)$

\Rightarrow S_{20} = \frac{20}{2} (2 \times 20 + (20 - 1) \times 1)

$\Rightarrow {S}_{20} = \cfrac{20}{2} \left(2 \times 20 + \left(20 - 1 \right) \times 1 \right)$

\Rightarrow S_{20} = 10 (40 + 19) = 10 \times 59 = 590

$\Rightarrow {S}_{20} = 10 \left(40 +19 \right) = 10 \times 59 = 590$

Hence correct answer is 590.

\frac{2}{3}, k, \frac{5}{8}

$\dfrac{2}{3},k, \dfrac{5}{8}$ are in AP, find the value of k.

0%
$\dfrac{48}{31}$
0%
$\dfrac{31}{48}$
0%
$\dfrac{17}{29}$
0%
$\dfrac{29}{17}$

Explanation

\frac{2}{3}, k, \frac{5}{8}

$\dfrac{2}{3},k, \dfrac{5}{8}$ are in AP.

Thus,

k - \frac{2}{3} = \frac{5}{8} - k

$k-\dfrac {2}{3} = \dfrac{5}{8}-k$

2 k = \frac{2}{3} + \frac{5}{8}

$2k=\dfrac{2}{3}+\dfrac{5}{8}$

2 k = \frac{31}{24}

$2k=\dfrac{31}{24}$

k = \frac{31}{48}

$k=\dfrac{31}{48}$ .

Find the number of terms of the A.P

- 12, - 9, - 6, . . . . ., 21

$-12,-9,-6,.....,21$ . If

1

$1$ is added to each term of this A.P., then find the sum of all terms of the A.P thus obtained.

0%
$10,66$
0%
$12,66$
0%
$14,66$
0%
None of these

Explanation

From given series, we have,

a_{n} = a + (n - 1) d

$a_n=a+(n-1)d$

21 = - 12 + (n - 1) 3

$21=-12+(n-1)3$

n = \frac{21 + 12}{3} + 1

$n=\dfrac{21+12}{3}+1$

∴ n = 12

$\therefore n=12$

upon adding

1

$1$ to the series, we get,

- 11, - 8, - 5, . . . . . . . . . . . . ., 22

$-11,-8,-5,.............,22$

S_{n} = \frac{n}{2} (a + l)

$S_n=\dfrac{n}{2}(a+l)$

S_{n} = \frac{12}{2} (- 11 + 22)

$S_n=\dfrac{12}{2}(-11+22)$

∴ S_{n} = 66

$\therefore S_n=66$

Find the sum of the first

15

$15$ terms of the sequence having

n

$n$ th term as

y_{n} = 9 - 5 n

${ y }_{ n }=9-5n\quad$

0%
$-390$
0%
$468$
0%
$-465$
0%
$-468$

Explanation

y_{n} = 9 - 5 n

$y_{n}=9-5n$

y_{1} = 9 - 5 = 4

$y_{1}=9-5=4$

y_{2} = 9 - 10 = - 1

$y_{2}=9-10=-1$

d = y_{2} - y_{1} = - 1 - 4 = - 5

$d=y_{2}-y_{1}=-1-4=-5$

Thus,

S_{15} = \frac{15}{2} [2 (4) + 14 (- 5)]

$S_{15}=\dfrac{15}{2}[2(4)+14(-5)]$

= \frac{15}{2} [8 - 70] = - 465

$=\dfrac{15}{2}[8-70]=-465$

Find the sum of the first

25

$25$ terms of an A.P whose

n

$n$ th term is given by

a_{n} = 2 - 3 n

${a}_{n}=2-3n$ .

0%
$-925$
0%
$-928$
0%
$-923$
0%
$-929$

Explanation

a_{n} = 2 - 3 n

$a_{n}=2-3n$

a_{1} = - 1

$a_{1}=-1$

a_{2} = - 4

$a_{2}=-4$

d = - 4 + 1 = - 3

$d=-4+1=-3$

S_{25} = \frac{25}{2} [- 2 + 24 (- 3)] = - 925

$S_{25}=\dfrac{25}{2}[-2+24(-3)]=-925$

Hence, option A is correct.

If two terms of an arithmetic progression are known, which values need to be determined to form the entire arithmetic progression?

0%
If $a$ and $d$ can be found out.
0%
If $a$ and $n$ can be found out.
0%
If $n$ and $d$ can be found out.
0%
If $t_n$ and $n$ can be found out.

Explanation

If two terms of an arithmetic progression, then the two terms can be represented using

t_{n} = a + (n_{1} - 1) d

$t_n = a + (n_1-1)d$ and

a + (n_{2} - 1) d

$a + (n_2-1)d$ . These two equations can be solved for finding the values of '

a

$a$ ' and '

d

$d$ '.

The arithmetic progression can be formed after the values of a and d are known.

The first term and the common difference of the arithmetic progression

3, 10, 17, 24, . . .

$3,10,17,24,...$ is?

0%
$7$ and $3$
0%
$3$ and $7$
0%
$3$ and $10$
0%
Not defined

Explanation

The first term of the arithmetic progression

3, 10, 17, 24, . . .

$3,10,17,24,...$ is

3

$3$ . The common difference or the difference between its consecutive terms is

t_{2} - t_{1} = 10 - 3 = 7

$t_2-t_1=10-3=7$ . Also,

t_{3} - t_{2} = 17 - 10 = 7

$t_3-t_2=17-10=7$ Thus the correct answer is option '

b

$b$ '.

The first term of an

A P

$AP$ whose

6^{t h}

$6^{th}$ terms is

5

$5$ and the

10^{t h}

$10^{th}$ terms is

9

$9$ is ____________.

0%
$4$
0%
$10$
0%
$0$
0%
$6$

Explanation

The gernal form of A.P is

a_{n} = a + (n - 1) d

$a_n=a+(n-1)d$

5 = a + (6 - 1) d = a + 5 d

$5=a+(6-1)d=a+5d$

9 = a + (10 - 1) d = a + 9 d

$9=a+(10-1)d=a+9d$

solving the two we get

$d=1$

$a+5*1=5$

$a=0$

An arithmetic progression is defined as a sequence that has a fixed _______ between its two consecutive numbers.

0%
difference
0%
ratio
0%
sum
0%
Cant say

Explanation

An arithmetic progression is a sequence of numbers that has a fixed difference between its two consecutive numbers. Thus the correct answer is option '

$a$ '.

If the sum of first

$2n$ terms of the A.P. 2,5,8,... is equal to the sum of first

$n$ terms of the A.P. 57, 59, 61,... ,then

$n$ equals

0%
10
0%
12
0%
11
0%
13

Explanation

Sum to n terms of the an A.P is

$S_n\;=\; \frac{n}{2}{(2a\;+\;(n\;-\;1)d)}$

Now equating sum of 2n terms of first AP to sum of n terms of second a.p we get;

$\frac{2n}{2}{(2a_1\;+\;(2n\;-\;1)d_1)}\;=\;\frac{n}{2}{(2a_2\;+\;(n\;-\;1)d_2)}$

$\frac{2n}{2}{(4\;+\;(2n\;-\;1)3)}\;=\;\frac{n}{2}{(114\;+\;(n\;-\;1)2)}$

On solving, we get

$n =11$

Which of the following are in arithmetic progression?

0%
$2,6,10,14,.....$
0%
$15,12,9,6,.....$
0%
$5,9,12,18,.....$
0%
$\dfrac{1}{2},\dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}.......$

Explanation

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.

Let us test the criteria for options given,

(A)

$2,6,10,14, . . . . .$

Here the difference of any two consecutive members is

$4$ . Hence this is an AP

(B)

$15,12,9,6, . . . . .$

Here the difference of any two consecutive members is

$-3$ . Hence this is an AP

(C)

$5,9,12,18, . . . . .$

Here the difference of any two consecutive members is not a constant

$(9-5=4, \ 12-9=3, \ 18-12=6)$ . Hence this is not an AP

(D)

$\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}, . . . . .$

Here the difference of any two consecutive members is not a constant

$(\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{-1}{6},\ \dfrac{1}{4}-\dfrac{1}{3}=\dfrac{-1}{12},\ \dfrac{1}{5}-\dfrac{1}{4}=\dfrac{-1}{20})$ . Hence this is not an AP

Correct options are (A) and (B).

What is the common difference in the arithmetic progression

$2,7,12,17,...$ ?

0%
$6$
0%
$5$
0%
$7$
0%
$9$

Explanation

The common difference in the arithmetic progression

$2,7,12,17,...$ is the difference between its two consecutive terms.

The difference between its two consecutive terms here is

$7-2=5$ , or

$12-7=5$ and so on.

Thus, the common difference is

$5$ .

If any one term of an arithmetic progression along with the position of the term in the A.P. and the common difference are known, then which of the following can be found out.

0%
The first term.
0%
The term before the known term.
0%
The term after the known term.
0%
All of the above.

Explanation

When one term of the arithmetic progression is known along with the common difference, then using that value of the common difference, the terms before and after the known term can be found out. Also substituting the values for

$d, n$ (the position of the term in the A.P. is known) and tn are known, then the value for a or the first term can be found out using the formula

$t_n = a + (n-1)d$ .

The sum of

$n$ terms of an A.P. is dependent on which of the following parameters?

0%
Only $a$ and $d$ .
0%
Only $n$ and $d$ .
0%
$a, n$ and $d$ .
0%
Only $a$ and $n$ .

Explanation

The sum of n terms of an arithmetic progression is dependent on the first term '

$a$ ', the common difference, d and the number of terms n as the formula is given as

$S_n$

$= \dfrac{n}{2} (2a+(n-1)d)$

Thus the correct answer is '

$c$ '.

$\sqrt{8},\sqrt{18},\sqrt{32}$ are in AP. Find the next term and common difference.

0%
$\sqrt {50}, \sqrt 2$
0%
$\sqrt {55}, 2$
0%
$\sqrt {48}, \sqrt 2$
0%
None

Explanation

$c.d.\>=\>\sqrt{18}-\sqrt8=3\sqrt2-2\sqrt2=\sqrt2\\\therefore\>next\>term\>=\>\sqrt{32}+\sqrt2\\=4\sqrt2+\sqrt2=5\sqrt2=\sqrt{50}$

Which term of the sequence 4, 9, 14, 19, ..., is 124 ?

0%
25
0%
30
0%
15
0%
35

Explanation

The given sequence is an A.P. with first term

$a=4$ and common difference

$d=5$ . Let 124 be the nth term of the given sequence. Then,

$a_n=124$

$a+(n-1)d=124$

$4+(n-1) \times 5 =124$

$n=25$

Hence, 25th term of the given sequence is 124.

Chose the correct option for the following sequence:

$-25, -23, -21, -19..$

0%
is an $A.P.$ Reason $d=3$
0%
is an $A.P.$ Reason $d=2$
0%
is an $A.P.$ Reason $d=4$
0%
is not an $A.P.$

Explanation

$-25,-23,-21,-19$

Common difference

$=2$

Hence series in AP

The common difference of the A.P.

$3, 5, 7,....$ is _________.

0%
$4$
0%
$2$
0%
$3$
0%
$-2$

Explanation

$d = a_2-a_1\\ = 5-3 = 2$ ,

where

$a_1$ &

$a_2$ are the first and second terms of the given sequence respectively.

Find the common difference of

$4,\dfrac{15}{2}, 11$

0%
$\frac { 7 }{ 2 }$
0%
$\frac { 2 }{ 7 }$
0%
$\frac { 11 }{ 41 }$
0%
$\frac { 41 }{ 11 }$

Explanation

The series is

$4,\dfrac{15}{2},11$

The common difference is

$\dfrac{15}{2}-4\\\dfrac{15-8}{2}=\dfrac 72$

The common difference of the A.P.: 3, 5, 7, ..... is __.

0%
$4$
0%
$2$
0%
$3$
0%
$-2$

Explanation

The given AP is

$3,\ 5,\ 7,\ .....$

To find out: The common difference of the A.P.

We know that, the common difference of an A.P. is the difference between any two consecutive terms of the A.P.

$\therefore \$ The common difference

$=5-3=2$ .

Hence, the common difference of the given A.P. is

$2$ .

A conference hall has nine rows of chairs which are in AP. If the sum of chairs of first five consecutive rows is

$230$ and

$7th$ row has

$52$ chairs, then the number of chairs in first row is

0%
$40$
0%
$42$
0%
$44$
0%
$46$

Explanation

Since, the number of chairs in the rows of a conference hall form an AP and the sum of chairs of five consecutive rows in a conference hall is 230

$\therefore (a-2d)+(a-d)+(a)+(a+d)+(a+2d) = 230$

$\implies 5a= 230 \implies a = 46$

Also, by given we have

$a_7 = 52 \implies a+6d = 52 \implies 46 +6d = 52 \implies 6d = 6 \implies d = 1$

Thus, the number of chairs in consecutive five rows are

$44, 45, 46, 47, 48, 48, 50, 51, 52$

So, the number of chairs in first row

$=44$ .

Riya arranged her magazines collection in three columns so that the number of magazines in columns form an

$AP$ . If the sum of number of magazines in all three columns is

$18$ and the number of magazines in last column is

$9,$ then first column has ____ number of magazines.

0%
$2$
0%
$3$
0%
$4$
0%
None of these

Explanation

Since, the number of magazines in columns form an AP

So by given, we have

$(a-d)+(a)+(a+d) = 18$

$\implies 3a = 18 \implies a =6$

Also, the number of magazines in last column

$=9$

$\implies (a+d) = 9 \implies 6+d = 9 \implies d = 3$

Thus, number of magazines in all three columns are

$3, 6, 9$

For an A.P.

$a_1,a_2,...a_n$ and

$a_1=\frac{5}{2},a_{10}=16$ .
If sum of

$n$ terms is

$110$ then

$n$ equals

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

$a_{1}=\dfrac{5}{2}$
Let the common difference be d.
Then

$a_{10}=a_{1}+9d$

$16=\dfrac{5}{2}+9d$

$\dfrac{27}{2}=9d$

$d=\dfrac{3}{2}$

Hence

$S_{n}=\dfrac{n}{2}(2(\dfrac{5}{2})+(n-1)\dfrac{3}{2})$

$=\dfrac{n}{4}(10+3(n-1))$

$=\dfrac{n}{4}(7+3n)$

Now

$S_{n}=110$
Hence

$\dfrac{n}{4}(7+3n)=110$

$n(7+3n)=440$

$3n^{2}+7n-440=0$

$n=11$ and

$n=\dfrac{-40}{3}$
Since

$n=\dfrac{-40}{3}$ is not possible, hence

$n=11$

$8^{th}$ term of an A.P is

$15$ , then the sum of

$15$ terms is

0%
$15$
0%
$0$
0%
$225$
0%
$\dfrac{225}2$

Explanation

$a_{8}=a_{1}+7d$

Hence

$a_{1}+7d=15$ ...(i)

$S_{15}=\dfrac{n}{2}[2a_{1}+(15-1)d]$

$S_{15}=\dfrac{15}{2}[2a_{1}+14d]$

$=15(a_{1}+7d)$

$=15(15)$ ...from i

$=225$ .

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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 6 - MCQExams.com

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