Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},...$ ​ are in A.P with ${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40$. Then find the sum of the first 15 terms of the given A.P.
(a) 200
(b) 280
(c) 120
(d) 150

Answer 1

In this question, We are given that ${{a}_{1}},{{a}_{2}},{{a}_{3}},...$ ​ are in A.P with ${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40$. Then there will be a common difference, say $d$ between the consecutive terms of the A.P. Now the formulate to calculate the ${{n}^{th}}$ terms of an A.P denotes by ${{a}_{n}}$ is given by
${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$. Also the sum of first $n$ terms say ${{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{n}}$ is given by $\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$. We will be using these formulas to calculate the sum of the first 15 terms of the given A.P.

Complete step by step answer:
We are given that ${{a}_{1}},{{a}_{2}},{{a}_{3}},...$ ​ are in A.P.
Also we are given that ${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40$.
Now since ${{a}_{1}},{{a}_{2}},{{a}_{3}},...$ ​ are in A.P there will be a common difference say $d$ between the consecutive terms of the A.P and the formulate to calculate the ${{n}^{th}}$ terms of an A.P denotes by ${{a}_{n}}$ is given by ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ .
Thus on substituting $n=1$ in ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$, we will have

& {{a}_{1}}={{a}_{1}}+\left( 1-1 \right)d \\\ & ={{a}_{1}} \end{aligned}$$ Now on substituting $$n=7$$ in $${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$$, we will have $$\begin{aligned} & {{a}_{7}}={{a}_{1}}+\left( 7-1 \right)d \\\ & ={{a}_{1}}+6d..........(1) \end{aligned}$$ Again on substituting $$n=16$$ in $${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$$, we will have $$\begin{aligned} & {{a}_{16}}={{a}_{1}}+\left( 16-1 \right)d \\\ & ={{a}_{1}}+15d.............(2) \end{aligned}$$ Now on substituting the values of equation (1) and equation (2) in the equation $${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40$$, we will have $${{a}_{1}}+\left( {{a}_{1}}+6d \right)+\left( {{a}_{1}}+15d \right)=40$$ On adding the common terms , we get $$3{{a}_{1}}+21d=40$$ Now by dividing the above expression by 3, we get $${{a}_{1}}+7d=\dfrac{40}{3}.............(3)$$ Since we know that the sum of first $$n$$ terms say $${{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{n}}$$ of an A.P is given by $$\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$$. Therefore the sum of first 15 terms of the given A.P, that is the sum of $${{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{15}}$$ is given by $${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=\dfrac{15}{2}\left( {{a}_{1}}+{{a}_{15}} \right)$$ Now on substituting $$n=15$$ in $${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$$, we will have $$\begin{aligned} & {{a}_{15}}={{a}_{1}}+\left( 15-1 \right)d \\\ & ={{a}_{1}}+14d..........(4) \end{aligned}$$ On substituting the value of equation (4) in the equation $${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{15}} \right)$$, we will have $$\begin{aligned} & {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=\dfrac{15}{2}\left( {{a}_{1}}+{{a}_{1}}+14 \right) \\\ & =\dfrac{15}{2}\left( 2{{a}_{1}}+14 \right) \\\ & =\dfrac{15}{2}\times 2\left( {{a}_{1}}+7 \right) \\\ & =15\left( {{a}_{1}}+7 \right) \end{aligned}$$ Now on substituting the value of equation (3) in the above expression, we will have $$\begin{aligned} & {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=15\left( \dfrac{40}{3} \right) \\\ & =5\times 40 \\\ & =200 \end{aligned}$$ Therefore the sum of the first 15 terms of the given A.P is equal to 200. **So, the correct answer is “Option A”.** **Note:** In this problem, we have to carefully use the formulas for the sum of first $$n$$ terms of an A.P and to find the $${{n}^{th}}$$ terms of an A.P in proper order. That is the formulate to calculate the $${{n}^{th}}$$ terms of an A.P denotes by $${{a}_{n}}$$ is given by $${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$$. Also the sum of first $$n$$ terms say $${{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{n}}$$ is given by $$\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$$.