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

Hint: 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
\[\begin{align}
  & {{a}_{1}}={{a}_{1}}+\left( 1-1 \right)d \\
 & ={{a}_{1}}
\end{align}\]
Now on substituting \[n=7\] in \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\], we will have
\[\begin{align}
  & {{a}_{7}}={{a}_{1}}+\left( 7-1 \right)d \\
 & ={{a}_{1}}+6d..........(1)
\end{align}\]
Again on substituting \[n=16\] in \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\], we will have
\[\begin{align}
  & {{a}_{16}}={{a}_{1}}+\left( 16-1 \right)d \\
 & ={{a}_{1}}+15d.............(2)
\end{align}\]
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
\[\begin{align}
  & {{a}_{15}}={{a}_{1}}+\left( 15-1 \right)d \\
 & ={{a}_{1}}+14d..........(4)
\end{align}\]
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
\[\begin{align}
  & {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=\dfrac{15}{2}\left( {{a}_{1}}+{{a}_{1}}+14 \right) \\
 & =\dfrac{15}{2}\left( 2{{a}_{1}}+14 \right) \\
 & =\dfrac{15}{2}\times 2\left( {{a}_{1}}+7 \right) \\
 & =15\left( {{a}_{1}}+7 \right)
\end{align}\]
Now on substituting the value of equation (3) in the above expression, we will have
\[\begin{align}
  & {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}=15\left( \dfrac{40}{3} \right) \\
 & =5\times 40 \\
 & =200
\end{align}\]
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 \[\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)\].