Question

The terms - 12, -9, -6 ...21 are in AP. Find the number terms in this AP. Now if 1 is added to each term of this AP, then find the sum of all the terms of the AP thus obtained.

Answer 1

Hint: Now we know that ${{n}^{th}}$ term of AP is given by $a + (n – 1)d. $ With the help of this formula, we can calculate the value of n. Now we will form new AP by adding 1 to each term. Now we know that sum of AP is given by the formula ${{S}_{n}}=\dfrac{n}{2}\times \left[ 2a+\left( n-1 \right)d \right]$ . Hence we can also find the sum of the new AP.

Complete step-by-step solution:
Now consider the given AP $- 12, -9, -6 ...21$
Now here the first term of AP is -12. Hence $a = -12.$
Here common difference is $ (-9) - (-12) = - 9 + 12 = 3. $
Hence the common difference d = 3.
And the last term of AP is 21. Hence ${{t}_{n}}=21$ .
Now we know that for an AP with first term a and common difference d, ${{n}^{th}}$ term is given by
$a + (n - 1)d $
Hence we have $21 = - 12 + (n - 1)(3)$
On opening the bracket we get $21 = –12 + 3n – 3.$
Adding and taking terms on LHS we get $21 + 15 = 3n$
Hence we have $36 = 3n.$
Now dividing the equation by 3 we get $n = 12.$
Hence there are 12 terms in AP.
Now let us form a new AP by adding 1 to each term.
Hence we get new AP as $-11, -8, -5…… 22$
Now since we are not changing the number of terms, the number of terms in the new AP will still be 12.
But in new AP we have $a = -11$ and $d =– 8 – (– 11) = – 8 + 11 = 3. $
Hence we have $a = - 11$ and $d = 3.$
Now we know that sum of n terms of AP is given by ${{S}_{n}}=\dfrac{n}{2}\times \left[ 2a+\left( n-1 \right)d \right]$
Hence using this we get ${{S}_{n}}=\dfrac{12}{2}\times \left[ 2\times \left( -11 \right)+\left( 12-1 \right)3 \right]$
Hence we have
$\begin{align}
  & {{S}_{n}}=6\times \left[ -22+33 \right] \\
 & \Rightarrow {{S}_{n}}=6\times 11 \\
 & \therefore {{S}_{n}}=66 \\
\end{align}$
Hence the sum of new AP is 66.

Note: Note that common difference can be calculated by subtracting any two consecutive terms. Now we call a sequence as AP if the common difference is the same. Hence when we add 1 to each term of AP we still get a common difference same and hence the new sequence is also an AP. But let us say you multiply 2 to each term of AP. then the difference between consecutive terms won’t be the same and hence the sequence obtained will not be an AP.