Question

Answer the following
The sum of tour consecutive terms which are in an arithmetic progression is 32 and the ratio of
the product of first and the last term to the product of two middle terms is 7.15. Find the
numbers​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

First term = p

Common Difference = d

${\boxed{\sf\:{Four\; Consecutive\;numbers}}}$

(p - 3d), (p - d), (p + d) and (p + 3d)

Hence,

p - 3d + p - d + p + d + p + 3d = 32

4p = 32

$\tt{\rightarrow p=\dfrac{32}{4}}$

p = 8 ......(1)

Now,

$\tt{\rightarrow\dfrac{(p-3d)(p+3d)}{(p-d)(p+d)}=\dfrac{7}{15}}$

15(p² - 9d²) = 7(p² - d²)

15p² - 135d² = 7p² - 7d²

15p² - 7p² = 135d² - 7d² 

8p² = 128d²

From Equation (1) :-

p = 8

8(8)² = 128d²

128d² = 512

$\tt{\rightarrow d^2=\dfrac{512}{128}}$

d² = 4

d = √4

d = 2

${\boxed{\sf\:{Four\; Consecutive\;numbers\;are:-}}}$

⇒p - 3d = 8 - (3 × 2) = 2

⇒p - d = 8 - 2 = 6

⇒p + d = 8 + 2 = 10

⇒p + 3d = 8 + (3 × 2) = 14

${\boxed{\sf\:{Four\; Consecutive\;numbers\;are:-}}}$

$\Large{\boxed{\sf\:{2,6,10\;and\;14}}}$

Answer 2

Solution:

Let the 4 consecutive terms of a.p are a, a + d, a + 2d and a + 3d.

According to question,

a + a + d + a + 2d + a + 3d = 32

=> 4a + 6d = 32    ...........(1)

Take 2 common from eq(1)

=> 2a + 3d = 16

$\sf{\implies a = \dfrac{16-3d}{2} \;\;\;\;.....(2)}$

Now, according to question.

$\sf{\implies \dfrac{a\times (a+3d)}{(a+d)\times(a+2d)}=\dfrac{7}{15}}$

$\sf{\implies 15a(a+3d)=7(a+d)(a+2d)}$

$\sf{\implies 15a^{2}+45ad=7(a^{2}+ad+2ad+2d^{2})}$

$\sf{\implies 15a^{2}+45ad=7a^{2}+21ad+14d^{2}}$

$\sf{\implies 8a^{2}+24ad-14d^{2}=0}$

By using splitting middle term method,

$\sf{\implies 8a^{2}+28ad-4ad-14d^{2}=0}$

$\sf{\implies 4a(2a+7d)-2d(2a+7d)=0}$

$\sf{\implies (4a-2d)(2a+7d)=0}$

$\sf{So,\;a=\dfrac{2d}{4}=\dfrac{d}{2} \;and\;a=-\dfrac{7d}{2}}$

$\sf{Put\;a=\dfrac{d}{2}\;in\;eq(2)}$

$\sf{\implies \dfrac{d}{2}=\dfrac{16-3d}{2}}$

$\sf{\implies d=16-13d}$

$\sf{\implies 4d = 16}$

$\sf{\implies d = 4}$

$\sf{Now,\;Put\;a=-\dfrac{7d}{2}\;in\;eq(2)}$

$\sf{\implies -\dfrac{7d}{2} = \dfrac{16-3d}{2}}$

$\sf{\implies -7d=16-3d}$

$\sf{\implies -4d=16}$

$\sf{\implies d = -4}$

$\sf{\implies d \neq -4,\;\;\because d = -4}$

$\sf{\implies a = -\dfrac{4}{2} = -2,\;\;a+d= -2-4=-6}$

And so on, the sum of all the terms cannot be negative,

So, $\sf{a \neq -\dfrac{7d}{2}}$

$\sf{\therefore d = 4}$

$\sf{\implies a = \dfrac{d}{2} = \dfrac{4}{2} = 2}$

and,

$\sf{\implies a + d = 2+4= 6}$

$\sf{\implies a+ 2d = 2+2\times 4 = 10}$

$\sf{\implies a+3d = 2+3\times 4=14}$

So, the numbers are 2, 6, 10 and 14.