Question

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P.

Answer 1

Hint : Apply the formula for the ${{n}^{th}}$ term of the AP, i.e., ${{T}_{n}}=a+(n-1)d$, where d is the common difference and a is the first term of the AP. Next form two equations from the information given to get the value of a and d. Then finally, to find the sum of first sixteen terms of the A.P., use the formula${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$, where $n=16$, a and d will be the values as obtained above.

Complete step by step solution :
In the question, we have to find the sum of the first sixteen terms of the A.P., where the sum of the third and the seventh terms of an A.P is 6 and their product is 8.
So, let a is the first term and d is the common difference of this AP. Now, use the ${{n}^{th}}$ term formula that is given as ${{T}_{n}}=a+(n-1)d$and find the third term ${{T}_{3}}$and the seventh term ${{T}_{7}}$.
So the third term is found as follows:

& \Rightarrow {{T}_{3}}=a+(3-1)d \\\ & \Rightarrow {{T}_{3}}=a+2d \\\ \end{aligned}$$ And the seventh term will be found as follows: $$\begin{aligned} & \Rightarrow {{T}_{7}}=a+(7-1)d \\\ & \Rightarrow {{T}_{7}}=a+6d \\\ \end{aligned}$$ Next, sum of the third and the seventh terms of an A.P is 6, so we have: $$\begin{aligned} & \Rightarrow {{T}_{3}}+{{T}_{7}}=6 \\\ & \Rightarrow a+2d+a+6d=6 \\\ & \Rightarrow a+4d=3\, \\\ & \Rightarrow a=3-4d\,\,\,\,\,\,\,\,\,\,(eq-1) \\\ \end{aligned}$$ Again, it is given that product of the third and the seventh terms of an A.P is 8, so we have: $$\begin{aligned} & \Rightarrow {{T}_{3}}\times {{T}_{7}}=8 \\\ & \Rightarrow \left( a+2d \right)\left( a+6d \right)=8\,\,\,\,\,\,\,\,\,\,(eq-2) \\\ \end{aligned}$$ So, putting equation 1 in equation 2, we get: $$\begin{aligned} & \Rightarrow \left( a+2d \right)\left( a+6d \right)=8\, \\\ & \Rightarrow \left( 3-4d+2d \right)\left( 3-4d+6d \right)=8\, \\\ & \Rightarrow \left( 3-2d \right)\left( 3+2d \right)=8\, \\\ & \Rightarrow \left( 9-4{{d}^{2}} \right)=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because (a-b)(a+b)={{a}^{2}}-{{b}^{2}} \\\ & \Rightarrow \left( 4{{d}^{2}} \right)=1\, \\\ & \Rightarrow d=\pm 0.5 \\\ \end{aligned}$$ So here we will have two series because there are two possible common differences and hence the first term will also differ. Now the first term of the two series is to be found. So when d=0.5, we have: $$\begin{aligned} & \Rightarrow a=3-4d\, \\\ & \Rightarrow a=3-4(0.5)\, \\\ & \Rightarrow a=1\, \\\ \end{aligned}$$ So here the first term of the first series will be 1 . Now when d=-0.5, we have: $$\begin{aligned} & \Rightarrow a=3-4d\, \\\ & \Rightarrow a=3-4(-0.5)\, \\\ & \Rightarrow a=5\, \\\ \end{aligned}$$ So here the first term of the second series will be 5. So when d=0.5, a=1 , this is series 1. Also, when d=-0.5, a=5, this is series 2. Now, our final task will be to find the sum of the first sixteen terms of these two series. The sum of the n terms of the series is given by the formula $${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$$. So for the first series where d=0.5 and a=1, the sum of first 16 terms will be given as: $$\begin{aligned} & \Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2(1)+(16-1)(0.5) \right] \\\ & \Rightarrow {{S}_{16}}=8\left[ 2+7.5 \right] \\\ & \Rightarrow {{S}_{16}}=76 \\\ \end{aligned}$$ Now, for the second series where d=-0.5 and a=5, the sum of first 16 terms will be given as: $$\begin{aligned} & \Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2(5)+(16-1)(-0.5) \right] \\\ & \Rightarrow {{S}_{16}}=8\left[ 10-7.5 \right] \\\ & \Rightarrow {{S}_{16}}=20 \\\ \end{aligned}$$ **Note** : The sum of the first n term of the AP can also be found using the formula$${{S}_{n}}=\dfrac{n}{2}\left[ first\,term+{{n}^{th}}\,term \right]$$. Also, make sure there is no calculation error while solving the equations of type $${{x}^{2}}=\sqrt{a}$$, as it will give $$x=\pm a$$.