Question

Let ${{a}_{1}},{{a}_{2}},......{{a}_{10}}$ be an AP and ${{a}_{5}}=27.$ What is the value of ${{a}_{10}}=?$
$\sum\limits_{i=1}^{30}{{{a}_{i}}}$ and $T=\sum\limits_{i=1}^{15}{a\left( 2i-1 \right)},S-2T=75.$
(a) 57
(b) 53
(c) 56
(d) 52

Answer 1

We will first find the sum of the first 30 terms in AP and then we will find the sum of the first 10 terms of the new AP which is made up to odd terms of our original AP by using the sum formula, i.e. ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right].$ The new AP will have the common difference as 2s. Once, we have S and T, we will solve S – 2T = 75. We will get, d = 5. Now, as we get d = 5, we will use ${{a}_{5}}=27.$ Using ${{a}_{n}}=a+\left( n-1 \right)d$ we will find the value of a, then again using this, we will find ${{a}_{10}}.$

Complete step-by-step solution:
We are given an AP ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},......$in which the fifth term is 27 and we have to find the tenth term. Let the first term of an AP be a and the common difference be d. Now the sum of the term is defined as $S=\sum\limits_{i=1}^{30}{{{a}_{i}}}.$
$S={{a}_{1}}+{{a}_{2}}+......+{{a}_{30}}$
We have the sum of n terms given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right].$ So, the sum of the 30 terms is given as,
${{S}_{30}}=\dfrac{30}{2}\left[ 2a+\left( 30-1 \right)d \right]$
$\Rightarrow {{S}_{30}}=15\left[ 2a+29d \right].....\left( i \right)$
Now T is defined as the sum of ${{a}_{2i-1}}$ terms, that is it is the sum of all the odd terms of the arithmetic progression.
$T={{a}_{1}}+{{a}_{3}}+{{a}_{5}}+.....+{{a}_{30}}$
The odd term of the AP with common difference d and the first term is ‘a’. Again forms an AP in which the first term is ‘a’ but the common difference is 2d.
So, we get, ${{a}_{1}}+{{a}_{3}}+{{a}_{5}}+.....+{{a}_{30}}$ is an AP with the first term ‘a’ and common difference ‘2d’. So,
T = Sum of such AP is given as T in summation till the fifteenth term. So,
${{T}_{15}}=\dfrac{15}{2}\left[ 2a+\left( 15-1 \right)2d \right]$
$\Rightarrow {{T}_{15}}=\dfrac{15}{2}\left[ 2a+28d \right]$
Simplifying further, we get,
$\Rightarrow {{T}_{15}}=15\left( a+14d \right)$
$\Rightarrow {{T}_{15}}=15a+210d......\left( ii \right)$
As, S – 2T is 75, so using (i) and (ii), we get,
$15\left( 2a+29d \right)-2\left( 15a+210d \right)=75$
Simplifying, we get,
$\Rightarrow 30a+435d-30a-420d=75$
Cancelling the like terms, we get,
$\Rightarrow 15d=75$
So, we get,
$\Rightarrow d=\dfrac{75}{15}=5$
Now, we get d = 5. As, ${{a}_{5}}=27$ and d = 5, we know ${{a}_{n}}$ is given as
${{a}_{n}}=a+\left( n-1 \right)d$
$\Rightarrow {{a}_{5}}=a+\left( 5-1 \right)d$
$\Rightarrow 27=a+4d$
$\Rightarrow a+4\times 5=27$
$\Rightarrow a=27-20$
$\Rightarrow a=7$
Now, as ${{a}_{n}}=a+\left( n-1 \right)d,$ so for ${{a}_{10}}$ we have a = 7, d = 5 and n = 10. So, we get,
${{a}_{10}}=7+\left( 10-1 \right)5$
$\Rightarrow {{a}_{10}}=7+45$
$\Rightarrow {{a}_{10}}=52$
So, we get the tenth term as 52.
Hence, the correct option is (d).

Note: Remember that ${{a}_{5}}$ is given as
${{a}_{5}}=a+\left( 5-1 \right)d$
$\Rightarrow {{a}_{5}}=a+4d$
Sometimes mistakes like ${{a}_{5}}=a+5d$ happens. So, always stay focussed when doing such calculations. T is defined as the sum of the odd terms of the AP. As we know,
${{a}_{1}}=a$
${{a}_{3}}=a+\left( 3-1 \right)d=a+2d$
${{a}_{5}}=a+\left( 5-1 \right)d=a+4d$
and so on.
They change by a common difference of 2d as the difference is the same, hence they form an AP.