Find the to n terms of the A-P- whose k-th- term is 5k-1-

Question

Toppr Admin · Accepted Answer

It is given that the kth term of the A-P- is 5k-1-Thus required sum is- -x2211-nk-1-5k-1-5-x2211-n1k-x2211-n11-5n-n-1-2-n-xA0- -xA0-by using-x2211-ni-1k-1-2-3-x2026-n-n-n-1-2-12n-5n-7-

Find the sum to $n$ terms of the $A . P .$ , whose $k^{t h}$ term is $5 k + 1$ .

It is given that the $k^{t h}$ term of the A.P. is $5 k + 1$ .

Thus required sum is, $= \sum_{k = 1}^{n} (5 k + 1) = 5 \sum_{1}^{n} k + \sum_{1}^{n} 1$

$= 5 \frac{n (n + 1)}{2} + n$ [by using $\sum_{i = 1}^{n} k = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$ ]

$= \frac{1}{2} n (5 n + 7)$

Find the sum to n terms of the A.P., whose kth term is 5k+1.

It is given that the kth term of the A.P. is 5k+1.Thus required sum is, =∑nk=1(5k+1)=5∑n1k+∑n11=5n(n+1)2+n [by using∑ni=1k=1+2+3+…+n=n(n+1)2]=12n(5n+7)

Find the sum to $n$ terms of the $A . P .$ , whose $k^{t h}$ term is $5 k + 1$ .

It is given that the $k^{t h}$ term of the A.P. is $5 k + 1$ .

Thus required sum is, $= \sum_{k = 1}^{n} (5 k + 1) = 5 \sum_{1}^{n} k + \sum_{1}^{n} 1$

$= 5 \frac{n (n + 1)}{2} + n$ [by using $\sum_{i = 1}^{n} k = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$ ]

$= \frac{1}{2} n (5 n + 7)$