Question

Find the sum of first 20 terms of an A.P, in which \[{{3}^{rd}}\] term is 7 and \[{{7}^{th}}\] term is two more than twice of 3 rd term.
\[\begin{align}
  & A.\text{ }700 \\
 & B.\text{ }720 \\
 & C.\text{ }740 \\
 & D.\text{ None of these} \\
\end{align}\]

Answer 1

Hint: We know that if a is the first term of an A.P and d is the common difference of the A.P, then \[{{n}^{th}}\] term of an A.P is equal to \[{{t}_{n}}\] where \[{{t}_{n}}=a+(n-1)d\]. Now we have to find the third term in terms of a and n. We know that the \[{{3}^{rd}}\] term is equal to 7. Now we have to find a relation between a and d. Let us assume this as equation (1). Now we have to find the seventh term in terms of a and n. We are also given that \[{{7}^{th}}\] term is two more than twice of \[{{3}^{rd}}\] term. Let us assume this equation as equation (2). By solving equation (1) and equation (2), we can get the value of a and n. We know that the sum of n terms of an A.P is equal to \[\dfrac{n}{2}[2a+(n-1)d]\]. We have to find the value of the sum of the first 20 terms. By using the above formula, we can find the value of the sum of 20 terms.

Complete step-by-step solution
We know that if a is the first term of an A.P and d is the common difference of the A.P, then \[{{n}^{th}}\] term of an A.P is equal to \[{{t}_{n}}\] where \[{{t}_{n}}=a+(n-1)d\].
Let us assume \[{{t}_{n}}=a+(n-1)d......(1)\]
To find the \[{{3}^{rd}}\] term we have to place the value of n is equal to 3 on equation (1).
\[\begin{align}
  & \Rightarrow {{t}_{3}}=a+(3-1)d \\
 & \Rightarrow {{t}_{3}}=a+2d......(2) \\
\end{align}\]
From the question, it was given that the \[{{3}^{rd}}\] term is equal to 7.
\[\Rightarrow a+2d=7.......(3)\]
To find the \[{{7}^{th}}\] term we have to place the value of n is equal to 7 on equation (1).
\[\begin{align}
  & \Rightarrow {{t}_{7}}=a+(7-1)d \\
 & \Rightarrow {{t}_{7}}=a+6d......(4) \\
\end{align}\]
We know that the \[{{7}^{th}}\] term is two more than twice of \[{{3}^{rd}}\] term.
\[\Rightarrow {{t}_{7}}=2{{t}_{3}}+2......(5)\]
Now we have to place equation (3) and equation (4) in equation (5), we get
\[\begin{align}
  & \Rightarrow a+6d=2(a+2d)+2 \\
 & \Rightarrow a+6d=2a+4d+2 \\
 & \Rightarrow a-2d=-2......(6) \\
\end{align}\]
Now we have to add equation (3) and equation (6), we get
\[\begin{align}
  & \Rightarrow a+2d+a-2d=7-2 \\
 & \Rightarrow 2a=5 \\
 & \Rightarrow a=\dfrac{5}{2}.....(7) \\
\end{align}\]
Now we have to substitute equation (7) in equation (3), we get
\[\begin{align}
  & \Rightarrow \dfrac{5}{2}+2d=7 \\
 & \Rightarrow 2d=7-\dfrac{5}{2} \\
 & \Rightarrow 2d=\dfrac{9}{2} \\
 & \Rightarrow d=\dfrac{9}{4}.....(8) \\
\end{align}\]
We know that the sum of n terms of an A.P is equal to \[\dfrac{n}{2}\left[ 2a+(n-1)d \right]\].
From the question, it is clear that we have to find the sum of the first 10 terms.
\[\Rightarrow n=20....(9)\]
From equation (7), equation (8) and equation (9), we get
\[\begin{align}
  & \Rightarrow Sum=\dfrac{20}{2}\left[ 2\left( \dfrac{5}{2} \right)+(20-1)\left( \dfrac{9}{4} \right) \right] \\
 & \Rightarrow Sum=10\left[ 5+(19)\left( \dfrac{9}{4} \right) \right] \\
 & \Rightarrow Sum=10\left( \dfrac{191}{4} \right) \\
 & \Rightarrow Sum=\dfrac{1910}{4} \\
 & \Rightarrow Sum=477.25.....(10) \\
\end{align}\]
From equation (10), it is clear that the sum of 20 terms is equal to 477.25,

Note: Some students have a misconception that if a is the first term of an A.P and d is the common difference of the A.P, then \[{{n}^{th}}\] term of an A.P is equal to \[{{t}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\]. If this misconception is followed, then the final answer may get interrupted. So, students should have a clear view of this concept.