Question

Sum of 4 consecutive numbers in an ap is 32 and the ratio of the first and last term to the product of two middle terms is 7:15.Feb 17, 2019

Answer 1

Solution: Let in such cases four numbers be a - 3d, a - d, a + d and a + 3d

→ (a - 3d) + (a - d) + (a + d) + (a - 3d) = 32

→ 4a = 32 → a = 8

Ratio: (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15

→ (a² - 9d²)/(a² - d²) = 7/15

→ 15a² - 135d² = 7a² - 7d²

→ 8a² = 128d²

→ 8 × 8²/128 = d²

→ 64/16 = d²

→ ± 2 = d

Hence A.P. is

• a - 3d = 8 - 3(2) = 2
• a - d = 8 - 2 = 6
• a + d = 8 + 2 = 10
• a + 3d = 8 + 3(2) = 14

OR

• a - 3d = 8 - 3(- 2) = 14
• a - d = 8 - (- 2) = 10
• a + d = 8 + (- 2) = 6
• a + 3d = 8 + 3(- 2) = 2

A.P. is 2, 6, 10, 14 or 14, 10, 6, 2

Answer 2

Answer:

• 4 consecutive terms are 2, 6, 10 and 14.

Step-by-step explanation:

Let 4 consecutive terms of A.P. are (a - 3d), (a + d), (a - d)  and (a + 3d).

Given:

• Sum of 4 consecutive terms of an A.P. = 32
• Ratio of the first and last term to product of two middle terms = 7:15

To Find:

• A.P.

Now, according to question,

Sum of 4 consecutive terms of an A.P. is 32. So,

⇒ (a - 3d) + (a + d) + (a - d) + (a + 3d) = 32

⇒ a - 3d + a + d + a - d + a + 3d = 32

⇒ 4a = 32

⇒ a = 32/4

⇒ a = 8

Now, it is given that the ratio of the first and last term to the product of two middle terms is 7:15.

So, according to question

$\longrightarrow \sf \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{7}{15}\\ \\ \\ {\underline {\bf Now, put\;the\;value\;of\;a\;in\;the\;above\;equation,}} \\ \\ \\ \longrightarrow \sf \dfrac{(8-3d)(8+3d)}{(8-d)(8+d)}=\dfrac{7}{15}\\ \\ \\ {\underline{\bf Now,\;we\;know\;that, (a-b)(a+b)=a^{2}-b^{2}}}.\\ \\ \\ \longrightarrow \sf \dfrac{64-9d^{2}}{64-d^{2}}=\dfrac{7}{15}\\ \\ \\ \longrightarrow \sf 15(64-9d^{2})=7(64-d^{2})\\ \\ \\ \longrightarrow \sf 960-135d^{2}=448-7d^{2}$

$\longrightarrow \sf 960-448=135d^{2}-7d^{2}\\ \\ \\ \longrightarrow \sf 512 = 128d^{2}\\ \\ \\ \longrightarrow \sf d^{2}=\dfrac{512}{128}\\ \\ \\ \longrightarrow \sf d^{2}=4\\ \\ \\ \longrightarrow \bf d=2$

So, four consecutive numbers are:

• (a - 3d) = 8 - 3 × 2 = 8 - 6 = 2
• (a - d) = 8 - 2 = 6
• (a + d) = 8 + 2 = 10
• (a + 3d) = 8 + 3 × 2 = 8 + 6 = 14

Hence, 4 consecutive terms are 2, 6, 10 and 14.

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