Question

Find the number of terms of the A.P -12, -9, -6, .....21. If 1 is added to each term of the A.P, then
find the sum of all terms of the A.P thus obtained.

Answer 1

Answer :

• The number of terms of the given A.P. = 12

• The sum of all terms of the A.P obtained after adding 1 to each term is 66

Step-by-step explanation :

Given :

• A.P. : -12, -9, -6, ....., 21

To find :

• the number of terms of the given A.P.
• the sum of all terms of the A.P., if 1 is added to each term of the given A.P

Solution :

➣ nth term of an A.P. is given by,

$\underline{\boxed{\bf a_n=a+(n-1)d}}$

where

a denotes first term

d denotes common difference

To find the number of terms of the given A.P., let nth term be 21 and

➛ first term, a = -12

➛ common difference, d = -9 - (-12) = -9+12 = 3

Substitute the values,

➙ aₙ = a + (n - 1)(3)

➙ 21 = -12 + 3n - 3

➙ 21 = 3n - 15

➙ 3n = 21 + 15

➙ 3n = 36

➙ n = 36/3

➙ n = 12

∴ The number of terms in the given A.P. = 12

Adding 1 to each term of the given A.P.,

-12 + 1 = -11

-9 + 1 = -8

-6 + 1 = -5

.........

21 + 1 = 22

So, the A.P. obtained is -11, -8 , -5 , ..... , 22

➛First term, a = -11

➛Common difference, d = -8 - (-11) = 11 - 8 = 3

Now, we have to find which term is 22 to get the value of n.

➙ aₙ = a + (n - 1)d

➙ 22 = -11 + (n - 1)(3)

➙ 22 = -11 + 3n - 3

➙ 22 = 3n - 14

➙ 3n = 22 + 14

➙ 3n = 36

➙  n = 36/3

➙  n = 12

➣ Sum of n terms is given by,

 $\underline{\boxed{\bf S_n=\dfrac{n}{2}[2a+(n-1)d]}}$

So, the sum of 12 terms of this A.P. is

 $\tt S_{12}=\dfrac{12}{2}[2(-11)+(12-1)(3)] \\\\ \tt S_{12}=6[-22+11(3)] \\\\ \tt S_{12}=6[-22+33] \\\\ \tt S_{12}=6[11] \\\\ \tt S_{12}=66$

∴ The sum of all terms of the A.P thus obtained is 66