Question

Let ${a_1},{a_2},{a_3},{a_4},\_\_\_\_\_,{a_{30}}$ be an A.P. $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $. If ${a_5} = 27$ and $S - 2T = 75$, then ${a_{10}}$ is equal to-
A) $57$
B) $47$
C) $42$
D) $52$

Answer 1

Hint:In this question we are given an A.P, so by using the formula for sum of the terms of the A.P
${S_n} = {a_1} + {a_2} + {a_3} + {a_4} + \_\_\_\_\_ + {a_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$,
Where $d$ is the common difference between the consecutive terms of A.P, we will reduce $S = \sum\limits_{i = 1}^{30} {{a_i}} $ in simpler form and then using the formula for $n{\text{th}}$ term of an A.P, ${a_n} = {a_1} + \left( {n - 1} \right)d$, we will reduce $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $ into simpler form. Further, we will substitute the values of $S{\text{ and }}T$ into given equation $S - 2T = 75$, and then again using the formula for $n{\text{th}}$ term of an A.P, ${a_n} = {a_1} + \left( {n - 1} \right)d$, we will solve ${a_5} = 27$.Now by solving these all equations we will get required information to find the value of ${a_{10}}$, using the formula for $n{\text{th}}$ term of an A.P, ${a_n} = {a_1} + \left( {n - 1} \right)d$.


Complete step-by-step answer:
We are given that-
${a_1},{a_2},{a_3},{a_4},\_\_\_\_\_,{a_{30}}$ is a A.P. ………………….(1)
Also sum of the all terms of the A.P. is given-
$S = \sum\limits_{i = 1}^{30} {{a_i}} $ ………………….(2)
Also, we are given that,
$T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $ ………………….(3)
We have the value of fifth term also,
${a_5} = 27$ ………………….(4)
Also, we are given a relation between $S{\text{ and }}T$
$S - 2T = 75$ ………………….(5)
And now we have to find the value of ${a_{10}}$.
Now in order to solve the above problem we will use the basic information about the A.P. or arithmetic progression-
We know if ${a_1},{a_2},{a_3},{a_4},\_\_\_\_\_,{a_n}$ are in A.P. then sum the $n$terms of a A.P. is equal to-
${S_n} = {a_1} + {a_2} + {a_3} + {a_4} + \_\_\_\_\_ + {a_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, ………………….(6)
Where $d$ is the common difference between the consecutive terms of A.P.
Now consider equation (2),
$S = \sum\limits_{i = 1}^{30} {{a_i}} $
Now expanding the summation, we get $d$ be the common difference of this A.P.
$S = {a_1} + {a_2} + {a_3} + {a_4} + \_\_\_\_\_ + {a_{30}}$, let
Now from (1), we know that ${a_1},{a_2},{a_3},{a_4},\_\_\_\_\_,{a_{30}}$ is a A.P. and using the formula (6), where $n = 30$, we get
$S = \dfrac{{30}}{2}\left( {2{a_1} + \left( {30 - 1} \right)d} \right)$
$ \Rightarrow S = 15\left( {2{a_1} + 29d} \right)$
$S = 30{a_1} + 435d$ ………………….(7)
Now let’s consider equation (3)
$T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $
Now expanding the summation, we get
$T = {a_{2\left( 1 \right) - 1}} + {a_{2\left( 2 \right) - 1}} + {a_{2\left( 3 \right) - 1}} + {a_{2\left( 4 \right) - 1}} + \_\_\_\_ + {a_{2\left( {15} \right) - 1}}$
Now solving further, we get
$T = {a_1} + {a_3} + {a_5} + {a_7} + \_\_\_\_ + {a_{29}}$ …………...(A)
Now we know if know if ${a_1},{a_2},{a_3},{a_4},\_\_\_\_\_,{a_n}$ are in A.P. then $n{\text{th}}$ term of a A.P. ${a_n}$ is ${a_n} = {a_1} + \left( {n - 1} \right)d$, ……………….(8)
Where $d$ is the common difference between the consecutive terms of A.P.
Now substituting (8), for $n = 3,5,7,9,\_\_\_\_,29$, in (A), we get
$T = {a_1} + {a_1} + 2d + {a_1} + 4d + {a_1} + 6d + \_\_\_\_ + {a_1} + 28d$
Now we get,
$T = 15{a_1} + \left( {2 + 4 + 6 + 8 + \_\_\_\_28} \right)d$
Now we can take $2$ common from second term,
$T = 15{a_1} + 2\left( {1 + 2 + 3 + 4 + \_\_\_\_ + 14} \right)d$ …………….(B)
Since we know that sum of first $n$ natural numbers is
$1 + 2 + 3 + 4 + \_\_\_\_ + n = \dfrac{{n\left( {n + 1} \right)}}{2}$
Now substituting this formula in (B), where $n = 14$, we get
\[T = 15{a_1} + 2\dfrac{{\left( {14\left( {15} \right)} \right)}}{2}d\]
Now solving further, we get
\[T = 15{a_1} + 210d\] ………………….(9)
Now we have obtained values of $S{\text{ and }}T$ and also, we have a given relation between $S{\text{ and }}T$ in equation (5)-
$S - 2T = 75$
So, now substituting values of $S{\text{ and }}T$ from (8) and (9), in (5)
$30{a_1} + 435d - 2\left( {15{a_1} + 210d} \right) = 75$
Now solving further, we get
$
  435d - 420d = 75 \\
   \Rightarrow 15d = 75$
So, $d = 5$ ………………….(10)
Now let’s consider the given equation (4)
${a_5} = 27$
Now using (8), the formula for $n{\text{th}}$ term of any A.P. ${a_n} = {a_1} + \left( {n - 1} \right)d$, where we take $n = 5$, we get
${a_1} + \left( {5 - 1} \right)d = 27$
Now substituting $d = 5$ from (10), we get
$
  {a_1} + \left( 4 \right)5 = 27 \\
   \Rightarrow {a_1} = 27 - 20 $
So, ${a_1} = 7$ ………………….(11)
Now according to the question, we have to find the value of term ${a_{10}}$,
So now again using (8), the formula for $n{\text{th}}$ term of any A.P. ${a_n} = {a_1} + \left( {n - 1} \right)d$, where we take $n = 10$, we get
${a_{10}} = {a_1} + \left( {10 - 1} \right)d$
Now substituting the value of $d{\text{ and }}{a_1}$, from (10) and (11), we get
${a_{10}} = 7 + \left( 9 \right)5$
 Now solving this we get
${a_{10}} = 52$

So, the correct answer is “Option D”.

Note:To solve this type of questions, where we have to find any term of the A.P, we will focus on finding the value of the first term of the A.P and the value of common difference of the consecutive terms of the A.P.Students should remember the definitions and formulas related to Arithmetic progressions for solving these types of questions.