Explanation
Identify which of the following list of numbers is an arithmetic progression?
Let a be the first term and d be the common difference.
The 4th term can be written as ‘a+3d’ and 7th term can be written as ‘a+6d’.
As per the question,
1. a+3d = 3a → 2a-3d=0
2. a+6d=2(a+2d)+1 →a+6d=2a+4d+1 →a-2d+1=0
Now solve the 2 quations, you will get:
d=2 & a=3
Hence, the first term is 3 and the common difference is 2.
The AP would be,
3,5,7,9,11,…..
Given, first term, $$a=10$$
last term, $$l=81$$
total number of terms, $$n=41$$
Sum, $$S_n=\dfrac{n}{2}[a+l]$$
$$=\dfrac{41}{2}[10+81]$$
$$=\dfrac{41}{2}\times91$$
$$=1865.5$$
Hence, the sum of the series is $$1865.5$$
$$ {\textbf{Step - 1: Find 12th term}} $$
$$ {\text{Given, a = 7 and d = 2}}{\text{.5}} $$
$$ {\text{a = first term and }} $$
$$ {\text{d = difference between two terms}} $$
$$ {\text{It is known that in an AP}}\;{{\text{n}}^{th}}{\text{ term will be}} $$
$$ {{\text{t}}_n}{\text{ = a + (n - 1)d}} $$
$$ {\text{Put the value of n as 12}}{\text{.}} $$
$$ \text{t}_{12} = 7 + (12 - 1)(2.5) $$
$$ { = 7 + 11 \times 2.5} $$
$$ { = 34.5} $$
$$ {\textbf{Hence, the correct answer is option B}}{\text{.}} $$
$$\\a=1,d=5\\\therefore\>a_n=a+(n-1)d$$
$$\\=1+(n-1)5\\=5n-4$$
Consider option (A),
$$\\4880=5n-4\\5n=4884\\n=(\dfrac{4884}{5})=not\>an\>integer$$
Consider option (B),
$$\\7881=5n-4\\5n=7885\\n=(\dfrac{7885}{5})=1577=integer$$
Consider option (C),
$$\\5890=5n-4\\5n=5894\\n=(\dfrac{5894}{5})=not\>an\>integer$$
Consider option (D),
$$\\7891=5n-4\\5n=7895\\n=(\dfrac{7895}{5})=1579=integer$$
$$\\\ast\>two\>options\>possible.$$
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