# Find the number of terms of the ap -12,-9,-6........12.if 1 is added to each term of this ap , then find the sum of all terms of the ap thus obtained

2
by nickii

Log in to add a comment

by nickii

Log in to add a comment

12=-12+(n-1)2

12= -12+2n-2

26 = 2n

n =13

∴ number of terms = 13

if 1 is added on both sides

series = -11,-8,.........13

sum of terms = n/2[2a+(n-1)d]

13/2[2(-11) + (13-1)2]

13/2[-22+24]

13/2[2]

13

A.P => -12, -9, -6,........,12.

first term = a = -12

2nd term = b = -9

last term = T = 12

no. of terms = n

so, common difference = d

d = b - a = (-9) - (-12)

or, d = -9 + 12 = 3

Now,

T = a + (n-1)d

or, 12 = -12 + (n - 1)3

or, (n - 1)3 = 12 + 12 = 24

or, n - 1 = 24/3 = 8

or, n = 8 + 1

n = 9

Total no. of terms = 9

and,sum of n terms(s) = n(a + T)/2

or, s = 9(-12 + 12)3 = 0

a/q,

Sum of n terms if 1 is added to every terms(S) = s + n

S = 0 + 9 = 9

Therefore sum of n terms if 1 is added to every terms is 9.