Question

First, second and seventh terms of an A.P. (all the terms are distinct), whose sum is 93, are in G.P. Fourth term of this G.P. is
A. 21
B. 31
C. 75
D. 375

Answer 1

This problem deals with both arithmetic and geometric progression. In an arithmetic progression, the consecutive terms differ by a common difference $d$ , with an initial term at the beginning. Whereas in the geometric progression, the consecutive terms differ by a common ratio $r$ , also with an initial term at the beginning.
The $n$ terms in an A.P. are generally expressed as:
$\Rightarrow a,\left( {a + d} \right),\left( {a + 2d} \right),........,\left( {a + (n - 1)d} \right).$
The $n$ terms in an G.P. are generally expressed as:
$\Rightarrow a,ar,a{r^2},........,a{r^{n - 1}}.$

Complete step-by-step answer:
The first term of the A.P. is given by:
$\Rightarrow {a_o} = a$
The second term of the A.P is:
$\Rightarrow {a_2} = a + (2 - 1)d$
$\Rightarrow {a_2} = a + d$
The seventh term of the A.P is:
$\Rightarrow {a_2} = a + (7 - 1)d$
$\Rightarrow {a_2} = a + 6d$
Given that the sum of the first, second and seventh terms of an A.P. is 93.
$\Rightarrow a + \left( {a + d} \right) + \left( {a + 6d} \right) = 93$
$\Rightarrow 3a + 7d = 93$
Given that these three terms are in G.P.
Which means that the ratio of the consecutive terms should be the same.
That is the terms in G.P. have a common ratio $r$ , as given by:
$\Rightarrow \dfrac{{a + d}}{a} = \dfrac{{a + 6d}}{{a + d}}$
$\Rightarrow {\left( {a + d} \right)^2} = a\left( {a + 6d} \right)$
$\Rightarrow {a^2} + {d^2} + 2ad = {a^2} + 6ad$
The ${a^2}$ term gets cancelled on both sides of the above equation, as given below:
$\Rightarrow {d^2} + 2ad = 6ad$
$\Rightarrow {d^2} - 4ad = 0$
$\Rightarrow d\left( {d - 4a} \right) = 0$
Here either $d = 0$ or $d = 4a$
As given that all the terms in the A.P are distinct, hence we cannot consider the expression $d = 0$ .
So considering the expression $d = 4a$ .
Substitute the expression of $d = 4a$ in the obtained expression $3a + 7d = 93$
$\Rightarrow 3a + 7d = 93$
$\Rightarrow 3a + 7\left( {4a} \right) = 93$
$\Rightarrow 3a + 28a = 93$
$\Rightarrow 31a = 93$
$\Rightarrow a = \dfrac{{93}}{{31}}$
$\therefore a = 3$
And hence finding the value of $d = 4a$ , as given below:
$\Rightarrow d = 4a$
$\Rightarrow d = 4(3)$
$\therefore d = 12$
So the first term, second term and seventh term in A.P are:
The first term is:
$\Rightarrow a = 3$
The second term is:
$\Rightarrow {a_2} = a + d$
$\Rightarrow {a_2} = 3 + 12$
$\Rightarrow {a_2} = 15$
The seventh term is:
$\Rightarrow {a_7} = a + 6d$
$\Rightarrow {a_7} = 3 + 6(12)$
$\Rightarrow {a_7} = 75$
So these first, second and seventh terms of an A.P are the first three terms in a G.P.
The first three terms in the G.P are 3, 15, 75.
Here the consecutive terms in a G.P. have a common ratio $r$ .
The common ratio is given by:
$\Rightarrow r = \dfrac{{15}}{3} = \dfrac{{75}}{{15}}$
$\therefore r = 5$
Here the common ratio $r$ of the G.P. is 5.
Hence the fourth term of the G.P. is given by:
$\Rightarrow {a_4} = 75(5)$
$\Rightarrow {a_4} = 375$

Final Answer: The fourth term of the G.P. is 375.

Note:
Please note that this problem can be done in another way also but with a slight change. Here we found the fourth term of the G.P by multiplying the previous term i.e, the third term with the common ratio $r$ . Instead of that we could rather substitute in the formula of general term in the G.P. That is the nth term in a G.P is given by $a{r^{n - 1}}$ , so to find the 4th term in the G.P, we could just substitute the values of $a,r$ and $n.$ So the 4th term is given by $a{r^{4 - 1}} = a{r^3}$