If the sum of the first 2n terms of the A.P. 2, 5, 8, ...., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ...., then n equals

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 11
$\frac{2 n}{2} [4 + (2 n - 1) 3] = \frac{n}{2} [2 \times 57 + (n - 1)^{2}]$
$2 (6 n + 1) = (144 + 2 n - 2)$
$2 (6 n + 1) = (2 n + 112)$
$12 n + 2 = 2 n + 112$
$10_{n} = 110$
$n = 11$

Suggest Corrections

Similar questions

Q. If the sum of the first 2n terms of the AP 2, 5, 8, ... is equal to the sum of the first n terms of A.P. 57, 59, 61, ... then what is the value of n?

Q. If the sum of the first

2 n

terms of

2, 5, 8, . . . .

is equal to the sum of the first

n

terms of

57, 59, 61....,

then

n

is equal to

Q. If the sum of first 2n terms of A.P . 2, 5, 8,... is equal to the sum of the first n terms of the A.P . 57, 59, 61, ....., then n equals-

Q. If the sum of the first 2n terms of the A.P. 2, 5, 8, ...., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ...., then n equals

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, . . .$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, . . .$ , then $n$ equals to

Join BYJU'S Learning Program

If the sum of the first 2n terms of the A.P. 2, 5, 8, ...., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ...., then n equals

The correct option is C 112n2[4+(2n−1)3]=n2[2×57+(n−1)2] 2(6n+1)=(144+2n−2) 2(6n+1)=(2n+112) 12n+2=2n+112 10n=110 n=11

The correct option is C 11
$\frac{2 n}{2} [4 + (2 n - 1) 3] = \frac{n}{2} [2 \times 57 + (n - 1)^{2}]$
$2 (6 n + 1) = (144 + 2 n - 2)$
$2 (6 n + 1) = (2 n + 112)$
$12 n + 2 = 2 n + 112$
$10_{n} = 110$
$n = 11$