Let 2 nd , 8 th and 44 th terms of a non-constant A.P. be respectively the 1 st , 2 nd and 3 rd terms of a G.P. If the first term of the A.P. is 1, then the sum of its first 20 terms is equal to [2024]

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Solution

Q.
Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. If the first term of the A.P. is 1, then the sum of its first 20 terms is equal to [2024]

1 990

2 980

3 970

4 960

Ans.
(3)

We have, first term of an A.P. =1.

Let the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. are $\frac{a}{r}, a, a r$ respectively.

Now, $T_{2} = \frac{a}{r} \Rightarrow 1 + (2 - 1) d = \frac{a}{r} \Rightarrow 1 + d = \frac{a}{r}$ ...(i)

$T_{8} = a \Rightarrow 1 + 7 d = a$ ...(ii)

$T_{44} = a r \Rightarrow 1 + 43 d = a r$ ...(iii)

Using equations (i) and (ii), we get

$\frac{1 + d}{1 + 7 d} = \frac{1}{r}$ ...(iv)

From (ii) and (iii), we get

$\frac{1 + 7 d}{1 + 43 d} = \frac{1}{r}$ ...(v)

$∴$ $\frac{1 + d}{1 + 7 d} = \frac{1 + 7 d}{1 + 43 d}$

$\Rightarrow 1 + 43 d + d + 43 d^{2} = 1 + 14 d + 49 d^{2}$

$\Rightarrow 6 d^{2} - 30 d = 0 \Rightarrow 6 d (d - 5) = 0 \Rightarrow d = 0, d = 5$

$∴ d = 5 (∵ d \neq 0)$

$∴ S_{20} = \frac{20}{2} [2 + 19 \times 5] = 10 [2 + 95] = 970$

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