Question

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

Answer 1

Answer:

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

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Answer 2

Answer:

The sum of n terms of the A.P. is $\frac{n}{2}[5n+7]$

Step-by-step explanation:

It is given that,

Kth term = 5k + 1

∴Tk = 5k + 1

Putting k = 1,

T1 = 5 x 1 +1

   = 5 + 1

   = 6

Similarly,

putting k = 2,

T2 = 11

putting k = 3,

T3 = 16...

and so on.

∴ Thus our A.P. is : 6, 11, 16, 21, 26, ......

Here, First term (a) = 6

Common Difference (d) = 11 - 6 = 5

As we know that, In an A.P. having 1st term a and common difference d, The Sum of n terms of the A.P. Sn = $\frac{n}{2}[2a+(n-1)d]$

∴$Sn=\frac{n}{2}[12+(n-1)*5]\\ =\frac{n}{2}[12+5n-5]\\ =\frac{n}{2}[7+5n]$

Hence, the sum of the n terms of the A.P. = $\frac{n}{2}[7+5n]$

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