Question

Let, ${t_n}$ be the ${n^{th}}$ term of an A.P. If $\sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r}}} = {10^{100}}$ and $\sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r - 1}}} = {10^{99}}$, then the common difference of A.P is
$\left( a \right)$ 1
$\left( b \right)$ 10
$\left( c \right)$ 9
$\left( d \right){10^{99}}$

Answer 1

Hint: In this particular question use the concept of arithmetic progression i.e. the succeeding term of the A.P is the sum of the preceding term and a constant value i.e. common difference, we can write this as ${a_2} = {a_1} + d$, so in general we can say that ${a_n} = {a_{n - 1}} + d$ , so use these concepts to reach the solution of the question.

Complete step-by-step answer:
$\sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r}}} = {10^{100}}$............... (1)
$\sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r - 1}}} = {10^{99}}$................ (2)
Now since the terms are in A.P.
Hence we can express the ${a_{2r}}^{th}$ term in terms of common difference of the A.P. and the ${a_{2r - 1}}^{th}$ term because the consecutive terms of an A.P. differ by the common difference, so expressing ${a_{2r}}^{th}$ term we get

Where d is the common difference of A.P.
Now apply $\sum\limits_{r = 1}^{{{10}^{99}}} {} $on both the sides of the above equation we have,
$ \Rightarrow \sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r}}} = \sum\limits_{r = 1}^{{{10}^{99}}} {\left( {{a_{2r - 1}} + d} \right)} $
Now using the splitting property of the summation i.e. $\sum {\left( {a + b} \right)} = \sum a + \sum b $ we have,
$ \Rightarrow \sum\limits_{r = 1}^{{{10}^{99}}} {{a_{2r}}} = \sum\limits_{r = 1}^{{{10}^{99}}} {\left( {{a_{2r - 1}} + d} \right)} = \sum\limits_{r = 1}^{{{10}^{99}}} {\left( {{a_{2r - 1}}} \right)} + \sum\limits_{r = 1}^{{{10}^{99}}} {\left( d \right)} $
Now substitute the values from equations (1) and (2) we have,
$ \Rightarrow {10^{100}} = {10^{99}} + \sum\limits_{r = 1}^{{{10}^{99}}} {\left( d \right)} $
Now simplify this we have,

$ \Rightarrow \sum\limits_{r = 1}^{{{10}^{99}}} {\left( d \right)} = {10^{100}} - {10^{99}}$
$ \Rightarrow \sum\limits_{r = 1}^{{{10}^{99}}} {\left( d \right)} = {10^{99}}\left( {10 - 1} \right) = 9{\left( {10} \right)^{99}}$
Now as we know that $\sum\limits_{r = 1}^n 1 = n$, so $\sum\limits_{r = 1}^{{{10}^{99}}} {\left( d \right)} = d{\left( {10} \right)^{99}}$ , so use this property in the above equation we have,
$ \Rightarrow d{\left( {10} \right)^{99}} = 9{\left( {10} \right)^{99}}$
Now simplify we have,
$ \Rightarrow d = 9$
So this is the required common difference of the A.P

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the property of arithmetic progression which s stated above, so using that property write the last term of the A.P in terms of second last term of A.P and common difference as above, then apply summation on both sides and splitting the summation as above then substitute the given values as above and simplify we will get the required value of common difference d.