Question

Find the value of 6 + 11 + 16 + 21 + ....... + 71.​

Answer 1

Given ,

First term = 6

Common difference = 11 - 6 = 5

Last term = 71

We know that , the general formula of an AP is given by

A = a + (n-1)d

So ,

71 = 6 + (n-1)5

65/5 = (n-1)

13 = n-1

n = 14

We know that , the sum of first n terms of an AP is given by

$\large \sf \fbox{s = \frac{n}{2}(a + l)}$

Substitute the known values , we get

Sum = 14/2(6 + 71)

Sum = 7(77)

Sum = 539

Hence , 6 + 11 + 16 + 21 + ....... + 71 is equal to 539

Answer 2

Answer:

a=6

d=5

an=71

an = a + (n-1) d

71 = 6 + (n-1) 5

71 - 6 / 5 = n-1

13 = n-1

n =14

sn = n/2 {2a + (n-1) d }

sn = 14/2 {2×6 + (14-1) 5}

sn = 7 (12 + 7×5)

sn = 7 (12 + 35)

sn = 7 (47)

sn = 329