The sum of the first ten terms of an A.P.

Question

The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8 . If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the ...

Accepted Answer

Let the first term of the A.P. be $a$ and common difference be $d$.

Then, from the question:

first term of the G.P. $=d$

common ratio of the G.P. $=a$

Now use the given conditions one by one.

1. Sum of first 10 terms of the A.P.
We know,

${S}_{10}=\frac{10}{2}[2a+(10-1)d]$

So,

$160=5(2a+9d)$

$2a+9d=32\, ...(1)$

2. Sum of first 2 terms of the G.P.
The G.P. has first term $d$ and common ratio $a$.

So its first two terms are:

$d,da$

Their sum is $8$:

$d+da=8$

$d(1+a)=8\, ...(2)$

3. Express $d$ from equation (1)
From

$2a+9d=32$

we get

$9d=32-2a$

$d=\frac{32-2a}{9}$

Substitute this in equation (2):

$\frac{32-2a}{9}(1+a)=8$

Multiply by $9$:

$(32-2a)(1+a)=72$

Expand:

$32+32a-2a-2{a}^{2}=72$

$32+30a-2{a}^{2}=72$

$-2{a}^{2}+30a-40=0$

Divide by $-2$:

${a}^{2}-15a+20=0$

4. Find values of $a$
Using quadratic formula:

$a=\frac{15\pm \sqrt{225-80}}{2}$

$a=\frac{15\pm \sqrt{145}}{2}$

Now we need the possible values of the first term of the G.P., which is $d$.

5. Find corresponding values of $d$
From equation (1):

$d=\frac{32-2a}{9}$

For the two values of $a$, the two values of $d$ are obtained. But the question asks for the sum of all possible values of the first term of the G.P., i.e. sum of both possible values of $d$.

Let the two roots of the quadratic in $a$ be ${a}_{1}$ and ${a}_{2}$.

Then corresponding values of $d$ are

${d}_{1}=\frac{32-2{a}_{1}}{9},\, {d}_{2}=\frac{32-2{a}_{2}}{9}$

So,

${d}_{1}+{d}_{2}=\frac{32-2{a}_{1}}{9}+\frac{32-2{a}_{2}}{9}$

${d}_{1}+{d}_{2}=\frac{64-2({a}_{1}+{a}_{2})}{9}$

For the quadratic

${a}^{2}-15a+20=0$

sum of roots is

${a}_{1}+{a}_{2}=15$

Hence,

${d}_{1}+{d}_{2}=\frac{64-2(15)}{9}$

${d}_{1}+{d}_{2}=\frac{64-30}{9}=\frac{34}{9}$

Therefore, the required sum is

$\frac{34}{9}$

So the correct option is A.

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