Question

14/TI 14The sum of five numbers in AP is 40. The product of the first and the last is 28. Find the numbers, DEL On sum of numbers in AP​

Answer 1

Step-by-step explanation:

★Correct question -

• The sum of five numbers in AP is 40. The product of the first and the last term is 28. Then find the numbers.

★Given -

• The sum of five numbers in AP is 40.
• The product of the first and the last term is 28.

★Solution -

Let the terms be (a - 2d), (a - d), (a), (a + d), (a + 2d).

According to the first statement,

$\rm\longmapsto \rm{ \bf a - 2d+ a - d + a + a + d + a + 2d = 40}$

$\rm\longmapsto \rm{ \bf5a = 40}$

$\rm\longmapsto \rm{ \bf \: a = \dfrac{ \cancel{40} \: \: ^{8} }{ {\cancel5} \: \: ^{1} } }$

$\rm\longmapsto \rm{ \bf \: \green{a = 8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)}$

Now, according to the second statement,

$\rm\longmapsto \rm{ \bf \: (a - 2d)(a + 2d) = 28}$

$\rm\longmapsto \rm{ \bf \: (8 - 2d)(8 + 2d) = 28 \: \: \: \: \: \{from \: (1) \}}$

$\rm\longmapsto \rm{ \bf \: {8}^{2} - {2d}^{2} = 28}$

$\rm\longmapsto \rm{ \bf \: 64- 4{d}^{2} = 28}$

$\rm\longmapsto \rm{ \bf \: - 4{d}^{2} = 28 - 64}$

$\rm\longmapsto \rm{ \bf \: - 4{d}^{2} = - 36}$

$\rm\longmapsto \rm{ \bf \: {d}^{2} = \dfrac{ {\cancel{- 36}} \: \: ^{9} }{{ \cancel - 4} \: \: ^{1} } }$

$\rm\longmapsto \rm{ \bf \: {d}^{2} = 9}$

$\rm\longmapsto \rm{ \bf \: \pink{{d}= ± \: 3}}$

Case 1

$\rm\longmapsto \rm{ \bf \: {{d}= 3}}$

Hence, A.P is, a - 2d = 8 - 2(3) = 8 - 6 = 2

a - d = 8 - 3 = 5

a = 8

a + d = 8 + 3 = 11

a + 2d = 8 + 2(3) = 8 + 6 = 14

Therefore, the A.P is 2, 5, 8, 11, 14...

Case 2

$\rm\longmapsto \rm{ \bf \: {{d}= - 3}}$

Hence, A.P is, a - 2d = 8 - 2( -3) = 8 + 6 = 14

a - d = 8 - ( -3) = 8 + 3 = 11

a = 8

a + d = 8 + ( -3) = 8 - 3 = 5

a + 2d = 8 + 2( -3) = 8 - 6 = 2

Therefore, the A.P is 14, 11, 8, 5, 2...