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Question: If \({{a}_{1}},{{a}_{2}},{{a}_{3}}........\) are in A.P. such that \({{a}_{1}}+{{a}_{7}}+{{a}_{16}}=...

If a1,a2,a3........{{a}_{1}},{{a}_{2}},{{a}_{3}}........ are in A.P. such that a1+a7+a16=40{{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40 , then the sum of the first 1515 terms of this A.P. is:
A. 200
B. 280
C. 120
D. 150

Explanation

Solution

We are given the arithmetic progression sequence, we will apply the conventional formula for nth{{n}^{th}} term that is , an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d and then we will find one equation in terms of a and d. Then we apply the formula for sum of n terms for an arithmetic progression which is Sn=n2(2a+(n−1)d){{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) and then we will put the value from the equation obtained earlier in terms of a and d.

Complete step by step answer:
Since it is given that the sequence a1,a2,a3,.......,an,....{{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n}},.... is in arithmetic progression, therefore it will be in the following form:
a,a+d,a+2d,.......,a+(n−1)d,.....a,a+d,a+2d,.......,a+\left( n-1 \right)d,.....
Where, an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d is the nth{{n}^{th}} term and aa is the first term in the sequence and dd is the common difference between terms.
Let’s take the first equation given in the equation that is: a1+a7+a16=40 .........Equation 1.{{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40\text{ }.........\text{Equation 1}\text{.} We will first find the values of these terms in terms of aa and dd :
We know that: an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d .
Therefore,
a1=a+(1−1)d⇒a1=a a7=a+(7−1)d⇒a7=a+6d a16=a+(16−1)d⇒a16=a+15d \begin{aligned} & {{a}_{1}}=a+\left( 1-1 \right)d\Rightarrow {{a}_{1}}=a \\\ & {{a}_{7}}=a+\left( 7-1 \right)d\Rightarrow {{a}_{7}}=a+6d \\\ & {{a}_{16}}=a+\left( 16-1 \right)d\Rightarrow {{a}_{16}}=a+15d \\\ \end{aligned}

Now we will put these values in equation 1:
a1+a7+a16=40 ⇒a+(a+6d)+(a+15d)=40 ⇒3a+21d=40⇒3(a+7d)=40 ⇒a+7d=403 ........... Equation 2. \begin{aligned} & {{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40 \\\ & \Rightarrow a+\left( a+6d \right)+\left( a+15d \right)=40 \\\ & \Rightarrow 3a+21d=40\Rightarrow 3\left( a+7d \right)\text{=40} \\\ & \Rightarrow a+7d=\dfrac{40}{3}\text{ }...........\text{ Equation 2}\text{.} \\\ \end{aligned}
Now we know that the sum of an arithmetic progression is: Sn=n2(2a+(n−1)d){{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) .
Now we have to find out the sum of first 15 terms of the A.P. , therefore, we will take n=15n=15

& \Rightarrow {{S}_{n}}=\dfrac{15}{2}\left( 2a+\left( 15-1 \right)d \right)=\dfrac{15}{2}\left( 2a+14d \right)=\dfrac{15}{2}\times 2\left( a+7d \right) \\\ & \Rightarrow {{S}_{n}}=15\left( a+7d \right) \\\ \end{aligned}$$ Now we will be putting the value of $a+7d$ from equation 3, therefore: $\begin{aligned} & {{S}_{n}}=15\times \dfrac{40}{3}=5\times 40=200 \\\ & \Rightarrow {{S}_{n}}=200 \\\ \end{aligned}$ Therefore, the sum of terms of the first 15 terms in the given arithmetic progression is $200$. **So, the correct answer is “Option A”.** **Note:** In questions like these, since only one condition is given you may not be able to find the values of $a$ and $d$ , as only one equation will be there. So when you find out the sum of the terms using the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ try and look out if you can fit the relation between $a$ and $d$ here, because in this way you will be able to get your answer.