IPMAT Indore 2022 - QA (SA) Q1 Explanation

Explanation:

(344)

It is given that the 3rd, 14th and 69th terms of given AP is forming three distinct and consecutive terms of a G.P. By observation, we can say that position 3rd, 14th and 69th follow a certain sequence i.e.,

(3 × 5)-1 = 14

(14 × 5)-1 = 69

Similarly, (69 × 5)-1 = 345-1= 344

So, next term in G.P. will be 344th term of the given AP.

Alternate Method

G1 = A3 = a +2d,

G2 = A14 = a +13d, and

G3 = A69 = a + 68d are in G.P.

We know that, if any three terms a, b, c are in G.P., then common ratio,

r = b a = c b or ac = b2

Applying the same,

(a + 2d) (a + 68d) = (a + 13d)^2

= a^2+68ad + 2ad + 136d^2 = a^2 + 26d + 169d^2

= 44ad = 33d^2

a/d = 3/4. Now let a = 3x and d = 4x

Common ratio, r = a+13d/a+2d

= 3x + 13 × 4x/3x+8x = 55x/11x = 5

Next term of GP will be G4 = G3 ×r = (a+68d) × 5 = (3x + 272x) × 5

= 275x × 5

= 1375 x

Let G4 be the nth term of given AP.

Therefore, a + (n - 1)d = 1375x

3x + (n - 1)4x = 1375x

n = 344.