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

Given :

The given A.P. is $-12, -9, -6,........21$.

To do :

We have to find the number of terms of an A.P., and the sum of all the terms if 1 is added to the given A.P.

Solution :

$-12, -9, -6,........21$.

Here, The first term $a=-12$, common difference $d=-9-(-12)=-9+12=3$, $a_{n}=21$.

$n^{th}$ term of an A.P. is given by, $a_{n}=a+(n-1)d$.

$21 = -12 + (n-1)3$

$21+12=(n-1)3$

$33=(n-1)3$

$\frac{33}{3}=n-1$

$n-1=11$

$n=11+1 = 12$.

Therefore, the number of terms in the given A.P. (n) is 12.

If 1 is added each term, then the A.P. becomes,

$-12+1, -9+1, -6+1,........21+1$.

The new A.P. is, $-11, -8, -5, ..........22$.

Now, first term $a=-11$, last term $l=22$, $d = -8-(-11)=-8+11=3$, $n=12$.

The sum of 'n' terms of an A.P is given by, $S_{n}=\frac{n}{2}[a+l]$.

$S_{n}=\frac{12}{2}[-11+22]$

$S_{n}=6[11]$

$S_{n}=66$.

Therefore, the sum of all terms of the new A.P. is 66.

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Updated on: 10-Oct-2022

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