JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q6

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MathematicsSingle correct+4 / -1Question 6 of 90

Consider two G.Ps. 2, $2^2$ , $2^3$ , ... and 4, $4^2$ , $4^3$ , ... of 60 and n terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}8}$ , then $\displaystyle\sum_{k=1}^nk(n-k)$ is equal to :

560

1540

1330

2600

Answer and worked solution

Official answer: C

GM. $\dfrac{1830+n(n+1)}{60+n}=\dfrac{225}8$ gives $n=20$ .

Sum $\sum k(20-k)=20\cdot210-2870=1330$ .

(C)

Topics:Geometric progression Special series sums

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 6