Question

a1 a2 a3 a4 a5 are in ap. Sum of the terms is equal to 30.product of the terms is equal to 3840. Find product of first and last term.?

Answer 1

AnswEr :

Let d be the common difference of the AP.

Consider the AP : a - 2d, a - d, a, a + d, a + 2d

According to the Question,

Sum of the five terms is 30.

$\longrightarrow \: \sf \: (a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 30 \\ \\ \longrightarrow \sf \: 5a = 30 \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: a = 6}}$

Product of the terms is 3840.

$\longrightarrow \sf (a - 2d)(a - d)(a)(a + d)(a - 2d) = 3840 \\ \\ \longrightarrow \sf \: a( {a}^{2} - {d}^{2} )( {a}^{2} - {4d}^{2} ) = 3840 \\ \\ \longrightarrow \sf \: (36 - {d}^{2} )(36 - {4d}^{2} ) = 640 \\ \\ \longrightarrow \sf \: 1296 - 144 {d}^{2} - {36d}^{2} + {4d}^{4} = 640 \\ \\ \longrightarrow \sf \: {4d}^{4} - {180d}^{2} + 656 = 0 \\ \\ \longrightarrow \sf \: {d}^{4} - 45 {d}^{2} + 164 = 0 \\ \\ \longrightarrow \sf \: {d}^{4} - {41d}^{2} - {4d}^{2} + 164 = 0 \\ \\ \longrightarrow \sf {d}^{2} ( {d}^{2} - 41) - 4( {d}^{2} - 41) = 0 \\ \\ \longrightarrow \sf \: ( {d}^{2} - 41)( {d}^{2} - 4 )= 0 \\ \\ \longrightarrow \boxed{ \boxed{ \sf d = \pm \sqrt{41} or \pm 2}}$

Therefore,

When d = ±2 and a = 6,

$\implies \sf \: (a - 2d)(a + 2d)] \\ \\ \implies \: \sf \: a^2 - 4d^2\\ \\ \implies \sf 20$

When d = ±√41 and a = 6,

$\implies \sf \: (a - 2d)( a + 2d) \\ \\ \implies \: \sf \: a^2 - 4d^2 \\ \\ \implies \sf \: -128$

Answer 2

✪GIVEN :- a1 a2 a3 a4 a5 are in ap, Sum of the terms is equal to 30& product of the terms is equal to 3840.

✪REQUIRED TO FIND :- Product of first and last term.

✍️SOLUTION:-

Let the AP be ,a-2d ,a-d ,a ,a+d ,a+2d.

» Sum of the terms = 30.

$\implies$ (a-2d )+( a-d )+ (a )+(a+d )+(a+2d). = 30.

$\implies$ 5a = 30.

$\implies \boxed{\: a = 6.}$

» Product of the terms = 3840.

$\implies \: ( a {}^{2} - 4d {}^{2} )a(a {}^{2} - d {}^{2} ) = 3840 \\ \\ \implies \: (36 - 4d {}^{2} )(6)(36 - d {}^{2} ) = 3840. \\ \\ \implies \: (9 - d {}^{2} )(36 - d {}^{2} ) = 160. \\ \\ \implies \: 324 - 45d {}^{2} + d {}^{4} = 160 \\ \\ \implies \: d {}^{4} - 45d {}^{2} + 164 = 0 \\ \\ \implies \: d {}^{2} (d {}^{2} - 41) - 4(d {}^{2} - 41) = 0 \\ \\ \implies \: d {}^{2} = 4. \\ \\ \boxed{ \implies \: d = + or - \: 2}$

Now,

$\implies \: a {}^{2} - 4d {}^{2} \: is \: the \: product. \\ \\ \implies \: 6 {}^{2} - 4 \times 4 = 36 - 16 = 20.$

Hence the answer is ,

$\implies \: \boxed{product \: = 20.}$

HOPE IT HELPS :)