Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

Let the three consecutive terms be a/r , a , ar.

acc to ques ,,

$\frac{a}{r} \times a \times ar \: = 512$

$→ {a}^{3} = {8}^{3}$

$→a = 8$

$the \: GP \: is \: \frac{8}{r} ,8,8r$

Now, 4 is added to first and second term and it form an AP :

$\frac{8}{r} + 4,8 + 4,8r \: are \: in \: AP$

$\frac{8}{r} + 4,12,8r \: are \: in \: AP$

so,

$→2 \times 12 = \frac{8}{r}+4 + 8r$

$→24-4 = \frac{8 + 8 {r}^{2} }{r}$

$→20r = 8 + 8 {r}^{2}$

$→8 {r}^{2} - 20r + 8 = 0$

$→2 {r}^{2} - 5r + 2 = 0$

$→2 {r}^{2} - 4r - r + 2 = 0$

$→2 r(r - 2) - (r - 2) = 0$

$→(2r - 1)(r - 2) = 0$

$→2r - 1 = 0 \: or \: r - 2 = 0$

$→r = \frac{1}{2} , \: 2$

Now, the original GP,

when, r = 1/2

$then, \: terms \: are \: \frac{8}{ \frac{1}{2} } ,8,8 \times \frac{1}{2} \\ = 16,8,4$

and, its sum = 16+8+4=28

again, r = 2

$then, \: terms \: are \: \frac{8}{2} ,8,8 \times 2 \\ = 4,8,16$

and,its sum = 4+8+16

hence, sum of original three terms of GP is 28.

Answer 2

EXPLANATION.

The product of three consecutive term of a G.P. = 512.

4 is added to each first and second term.

The three terms now form an A.P.

As we know that,

Products of three consecutive term of G.P.

⇒ a/r x a x ar = 512.

⇒ a x a x a = 512.

⇒ a³ = 512.

⇒ a³ = 8 x 8 x 8.

⇒ a³ = 8³.

⇒ a = 8.

4 is added to each first and second term.

⇒ a/r + 4, a + 4, ar. - - - - - (A.P).

⇒ 8/r + 4, 8 + 4, 8r.

⇒ 8/r + 4, 12, 8r.

As we know that,

Conditions of an A.P.

⇒ 2b = a + c.

Put the value in the equation, we get.

⇒ 2(12) = 8/r + 4 + 8r.

⇒ 24 = 8 + 4r + 8r²/r.

⇒ 24r = 8 + 4r + 8r².

⇒ 8r² + 4r + 8 - 24r = 0.

⇒ 8r² - 20r + 8 = 0.

Factorizes the equation into middle term splits, we get.

⇒ 8r² - 16r - 4r + 8 = 0.

⇒ 8r(r - 2) - 4(r - 2) = 0.

⇒ (8r - 4)(r - 2) = 0.

⇒ r = 1/2  and  r = 2.

Three terms are.

⇒ a/r, a, ar.

If r = 1/2 and a = 8.

⇒ 8/1/2, 8, 8 x 1/2.

⇒ 16, 8, 4.

If r = 2 and a = 8.

⇒ 8/2, 8, 8 x 2.

⇒ 4, 8, 16.

Sum of three numbers,

16 + 8 + 4 = 28.

Option [B] is correct answer.