the sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first - Maths - Arithmetic Progressions

Question

the sum of four consecutive numbers in an AP is 32 and the ratio
of the product of the first and the last terms to the product of
the two middle terms is 7:15 find the nos? plz its urgent

Priyanka Kedia · Accepted Answer

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

Given, the sum of four consecutive numbers of an A.P. is 32. Therefore,

$a + (a + d) + (a + 2 d) + (a + 3 d) = 32 \Rightarrow 4 a + 6 d = 32 \Rightarrow 2 a + 3 d = 16 \Rightarrow a = \frac{16 - 3 d}{2} ....... (1)$

And, ratio of the product of first and last terms to the product of two middle terms is 7 : 15. Therefore,

$\frac{a (a + 3 d)}{(a + d) (a + 2 d)} = \frac{7}{15} \Rightarrow 15 a (a + 3 d) = 7 (a + d) (a + 2 d) \Rightarrow 15 a^{2} + 45 a d = 7 a^{2} + 21 a d + 14 d^{2} \Rightarrow 8 a^{2} + 24 a d - 14 d^{2} = 0 \Rightarrow 4 a^{2} + 12 a d - 7 d^{2} = 0 \Rightarrow 4 a^{2} + 14 a d - 2 a d - 7 d^{2} = 0 \Rightarrow 2 a (2 a + 7 d) - d (2 a + 7 d) = 0 \Rightarrow (2 a + 7 d) (2 a - d) = 0 \Rightarrow a = - \frac{7 d}{2} a n d a = \frac{d}{2} ....... (2) S u b s t i t u t e v a l u e o f a f r o m e q u a t i o n (1) in e q u a t i o n (2), w e h a v e, w h e n a = - \frac{7 d}{2}, - \frac{7 d}{2} = \frac{16 - 3 d}{2} \Rightarrow - 7 d = 16 - 3 d \Rightarrow d = - 4 S o, a = - \frac{7 \times (- 4)}{2} = 14 \Rightarrow a = 14 T h u s, n u m b e r a r e 14, (14 - 4), (14 - 2 \times 4) a n d (14 - 3 \times 4) i . e, 14, 10, 6 a n d 2. A n d, w h e n a = \frac{d}{2}, \frac{d}{2} = \frac{16 - 3 d}{2} \Rightarrow d = 16 - 3 d \Rightarrow d = 4 S o, a = \frac{(4)}{2} = 2 \Rightarrow a = 2 T h u s, n u m b e r a r e 2, (2 + 4), (2 + 2 \times 4) a n d (2 + 3 \times 4) i . e, 2, 6, 10 a n d 14.$

the sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15.find the nos? plz its urgent.