Question

If for an A.P. $\left\{{\mathrm{a}}_{\mathrm{n}}\right\},{\mathrm{a}}_1+{\mathrm{a}}_5+{\mathrm{a}}_{15}+{\mathrm{a}}_{26}+{\mathrm{a}}_{36}+{\mathrm{a}}_{40}=210$ then ${\mathrm{S}}_{40}$

• A. $2100$
• B. $700$
• C. $1400$
• D. None of these

Answer 1

The correct option is C

$1400$

Explanation for the correct answer:

Step 1: Information required for the solution

To find the last term of an A.P, we use the formula,

${\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{d}$

where,

${\mathrm{a}}_{\mathrm{n}}=$ the n^th term in the sequence

${\mathrm{a}}_1=$the first term in the sequence

$\mathrm{d}=$the common difference between terms

Since ${\mathrm{a}}_1+{\mathrm{a}}_5+{\mathrm{a}}_{15}+{\mathrm{a}}_{26}+{\mathrm{a}}_{36}+{\mathrm{a}}_{40}=210$ which can be rewritten as

$\left(\mathrm{a}+\left(1-1\right)\mathrm{d}\right)+\left(\mathrm{a}+\left(5-1\right)\mathrm{d}\right)+\left(\mathrm{a}+\left(15-1\right)\mathrm{d}\right)+\left(\mathrm{a}+\left(26-1\right)\mathrm{d}\right)+\left(\mathrm{a}+\left(36-1\right)\mathrm{d}\right)+\left(\mathrm{a}+\left(40-1\right)\mathrm{d}\right)=210$

After further simplification we get,

$\begin{array}{rcl}\therefore a+a+4d+a+14d+a+25d+a+35d+a+39d& =& 210\\ & \Rightarrow & 6a+117d=210\\ & \Rightarrow & 2a+39d=70\dots \left(1\right)\end{array}$

Step 2: Calculation of the sum of the A.P

The formula used for the calculation of the sum of an A.P is

$\begin{array}{rcl}{\mathbf{s}}_{\mathit{n}}& \mathbf{=}& \mathbf{\left(}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{\right)}\mathbf{(}\mathbf{\text{first term}}\mathbf{+}\mathbf{\text{last term}}\mathbf{)}\end{array}$

From the A.P, we know that $n=40,\text{First term}=a,\text{and}\text{Last term}=a+39d$, then

$\begin{array}{rcl}{S}_{40}& =& \frac{40}{2}(\mathrm{a}+\mathrm{a}+39\mathrm{d})\\ & =& 20(2\mathrm{a}+39\mathrm{d})\\ & =& 20\left(70\right)\left[\text{from equation}\left(1\right)\right]\\ & =& 1400\end{array}$

Hence, the correct option is (C).