Question

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},...,{{a}_{n}}\]are in A.P and \[{{a}_{1}}+{{a}_{4}}+{{a}_{7}},...,{{a}_{16}}=114\] then \[{{a}_{1}}+{{a}_{6}}+{{a}_{11}}+{{a}_{16}}\]is equal to:
A.38
B. 98
C. 76
D. 64

Answer 1

Hint: Find the sum of an A.P series. To find the sum of the A.P terms we will use the conventional formula and calculate the sum of an A.P series.
The given expression is \[{{a}_{1}}+{{a}_{4}}+{{a}_{7}}+{{a}_{10}}+{{a}_{13}}+{{a}_{16}}=114\].
Formulas used:
The formula for sum of an A.P.
\[\begin{align}
  & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ S}=\dfrac{n}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{n}} \right) \\
 & {{\text{a}}_{n}}=\text{ the last term in the sequence} \\
 & {{a}_{1}}=\text{ the first term in the sequence} \\
 & n=\text{ the number of terms in the sequence} \\
 & S=\text{ Sum of the series} \\
\end{align}\]

Complete step-by-step answer:
First step will be implementing the series in the equation.
the value of first and last terms are obtained.
We will use the value of first and last terms in the given expression to calculate the value of an A.P series.
\[\begin{align}
  & {{a}_{1}}+{{a}_{4}}+{{a}_{7}}+{{a}_{10}}+{{a}_{13}}+{{a}_{16}}=114 \\
 & \therefore \text{ }\!\!~\!\!\text{ S}=\dfrac{n}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{n}} \right) \\
 & 114=\dfrac{6}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{16}} \right) \\
 & \left( {{a}_{1}}+{{\text{a}}_{16}} \right)=\dfrac{2}{6}\cdot 114 \\
 & \left( {{a}_{1}}+{{\text{a}}_{16}} \right)=38 \\
 & \text{We have, }{{a}_{1}}+{{a}_{6}}+{{a}_{11}}+{{a}_{16}} \\
 & \text{S}=\dfrac{n}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{n}} \right) \\
 & S=\dfrac{4}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{16}} \right) \\
 & S=\dfrac{4}{2}\cdot 38\text{ }\left[ \because \left( {{a}_{1}}+{{\text{a}}_{16}} \right)=38 \right] \\
 & S=76 \\
\end{align}\]
Thus, the sum of \[{{a}_{1}}+{{a}_{6}}+{{a}_{11}}+{{a}_{16}}\] is 76
So, the correct answer is “Option C”.


Note: A mathematical sequence in which the difference between two consecutive terms is always constant and it is called Arithmetic Progression.
The finite portion of an A.P is known as A.P and therefore the sum of finite A.P is known as Arithmetic series.
The formula for sum of A.P series.
\[\begin{align}
  & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ S}=\dfrac{n}{2}\cdot \left( {{a}_{1}}+{{\text{a}}_{n}} \right) \\
 & {{\text{a}}_{n}}=\text{ the last term in the sequence} \\
 & {{a}_{1}}=\text{ the first term in the sequence} \\
 & n=\text{ the number of terms in the sequence} \\
 & S=\text{ Sum of the series} \\
\end{align}\]