Question

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 ?

Answer 1

Given AP is − 12, − 9, − 6, ....., 12

The first term of the AP is − 12 and the common difference is 3.

Let 12 be the n^th term of the AP.

Now, a = −12; d = 3; a_n = 12

a_n = a + (n − 1)d

Therefore, 12 = − 12 + (n − 1) × 3

⇒ n = 9

Sum of the terms of the AP `n/2[2a+(n-1)d]=9/2[2(-12)+(9-1)3]=0`

If 1 is added to each term of the AP, the sum of all the terms of the new AP will increase by n, i.e. 9.

Therefore, the sum of all the terms of the new AP will be 0 + 9 = 9.

Answer 2

First term, 

\[a_1 = - 12\]  Common difference,

\[d = a_2 - a_1 = - 9 - \left( - 12 \right) = 3\] 

\[a_n = 21\]
\[ \Rightarrow a + \left( n - 1 \right)d = 21\]
\[ \Rightarrow - 12 + \left( n - 1 \right) \times 3 = 21\]
\[ \Rightarrow 3n = 36\]
\[ \Rightarrow n = 12\]

Therefore, number of terms in the given A.P. is 12.
Now, when 1 is added to each of the 12 terms, the sum will increase by 12.
So, the sum of all terms of the A.P. thus obtained

\[= S_{12} + 12\]
\[ = \frac{12}{2}\left[ 2\left( - 12 \right) + 11\left( 3 \right) \right] + 12\]
\[ = 6 \times \left( 9 \right) + 12\]
\[ = 66\]