Question

find the sum to n terms of AP whose kth term is 5K + 1

Answer 1

On putting k=1,we get 1st term a1=5(1)+1=6,a2=5(2)+1=11, common difference d=a2-a1=11-6=5
So sum of n terms =n/2[2(6)+(n-1)5]
=n/2(12+5n-5)
=n/2(5n+7)
=(5n^2+7n)/2

Answer 2

we have to find the sum of n terms of an AP of which kth term is 5k + 1

Solution : method 1 :

It has given that, $t_k=5k+1$

sum of n terms = $\Sigma^{k=n}_{k=1}t_k$

= $\Sigma^{k=n}_{k=1}(5k+1)$

=$\Sigma^{k=n}_{k=1}(5k)+\Sigma^{k=n}_{k=1}(1)$

= 5n(n + 1)/2 + n

= n [5n/2 + 5/2 + 1]

= n [ 5n/2 + 7/2 ]

= n (5n + 7)/2

Therefore the sum of n terms of AP is n(5n + 7)/2.

method 2 : kth term, $t_k=(5k+1)$

first term, t₁ = 5 × 1 + 1 = 6

2nd term, t₂ = 5 × 2 + 1 = 11

common difference, d = 11 - 6 = 5

using formula, $S_n=\frac{n}{2}(2t_1+(n-1)d)$

so, sum of n terms = n/2 [2 × 6 + (n - 1) × 5]

= n/2 [12 + 5n - 5 ]

= n(5n + 7)/2

Therefore the sum of n terms of AP is n(5n + 7)/2