Question

Sun of the first 4 term of an ap is 72 sum of the first 9 term is alsi 72 wht us the 5 term og the sequence

Answer 1

Questión : The sum of the first 4 terms of an ap is 72. The sum of the first 9 terms is also 72. What is the 5th term of the sequence?

Solution :

• The sum of the first 4 terms of the ap is equal to the sum of the first 9 terms ( = 72)

For an ap ;

Sₙ = ½n [ 2a + (n-1) d]

S₄ = ½ × 4 [ 2a + 3d ] = 72

> 2( 2a + 3d) = 72

> 2a + 3d = 36

S₉ = ½ × 9 [ 2a + 8d] = 72

> 9( 2a + 8d) = 144

> 2a + 8d = 16

Subtracting the second equation from the first :

> 3a - 8d = 36 - 16

> -5d = 20

> d = -4

2a + 3(-4) = 36

> 2a - 12 = 36

> 2a = 48

> a = 24

The second term is 20, third is 16 and so on ..

Answer : The sequence satisfying this criteria is 24, 20, 16, 12, ....

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Answer 2

Question : Sum of the first 4 terms of an A.P is 72 & sum of the first 9 terms of an A.P is also 72 . What is the 5 th term of the sequence ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀▪︎⠀⠀We know that , Formula to calculate sum of an A.P and that's given by :

$\\\qquad \star\:\: \underline {\boxed {\pmb{\sf S_n = \dfrac{n}{2}\bigg\lgroup 2a + ( n - 1 ) d \bigg\rgroup}}}$

⠀⠀⠀⠀⠀⠀⠀⠀Here , n is the n th Term of an A.P & d is the Common Difference of an A.P .

⠀⠀⠀⠀CASE I : The sum of first 4 terms of an A.P is 72 .

$\\\qquad:\implies \sf S_n = \dfrac{n}{2} \bigg\lgroup 2a + ( n - 1 ) d \bigg\rgroup \\\\\\ \qquad:\implies \sf 72 = \dfrac{4}{2}\bigg\lgroup 2a + ( 4 - 1 ) d \bigg\rgroup\\\\\\ \qquad:\implies \sf 72 = 2\bigg\lgroup 2a + 3d \bigg\rgroup\\\\\\\qquad:\implies \sf 36 = 2a + 3d \qquad \bigg\lgroup Eq^n \:I \bigg\rgroup$

⠀⠀⠀⠀CASE II : The sum of first 9 terms of an A.P is 72 .

$\qquad:\implies \sf S_n = \dfrac{n}{2} \bigg\lgroup 2a + ( n - 1 ) d \bigg\rgroup \\\\\\ \qquad:\implies \sf 72 = \dfrac{9}{2}\bigg\lgroup 2a + ( 9 - 1 ) d \bigg\rgroup\\\\\\ \qquad:\implies \sf 72 \times 2 = 9 \bigg\lgroup 2a + 8d \bigg\rgroup \\\\\\\qquad:\implies \sf 144 = 9 ( 2a + 8d ) \\\\\\\qquad:\implies \sf 16 = 2a + 8d \bigg\lgroup Eq^n \:II \bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Subtracting \: Eq^n \: II \: from \: Eq^n \: I \::}}$

$\qquad \dashrightarrow \sf ( 36 - 16 ) = \bigg( \{ 2a - 2a \} + \{ 8d - 3d \} \bigg)\\\\\\ \qquad \dashrightarrow \sf 16 = \bigg( \{ 3d - 8d \} \bigg) \\\\\\ \qquad \dashrightarrow \sf -5d = 20 \\\\ \\ \qquad \dashrightarrow \sf \underline {\boxed{\pmb{\frak{ d \: ( \: or \: Common \: Difference \:)\:=\; –4 \:}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\: of \: d \: in \: Eq^n \: II \::}}$

$\qquad \dashrightarrow \sf 16 = 2a + 8d \bigg\lgroup Eq^n \:II \bigg\rgroup \\\\\\ \qquad \dashrightarrow \sf 16 = 2a + 8d \\\\\\ \qquad \dashrightarrow \sf 16 = 2a + 8(-4) \\\\ \qquad \dashrightarrow \sf 16 = 2a - 32 \\\\\\ \qquad \dashrightarrow \sf 16 + 32 = 2a \\\\\\ \qquad \dashrightarrow \sf 48 = 2a \\\\\\ \qquad \dashrightarrow \sf \underline {\boxed{\pmb{\frak{ a \: ( \: or \: First \: Term \:)\:=\; 24 \:}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding 5 th Term of an A.P :

⠀⠀▪︎⠀⠀We know that , Formula to calculate n th Term of an A.P and that's given by :

$\\\qquad \star \underline {\boxed {\pmb{\sf a_n = a + \bigg\lgroup n - 1 \bigg\rgroup d }}}$

⠀⠀⠀⠀⠀⠀⠀⠀Here , n is the n th Term of an A.P & d is the Common Difference of an A.P .

$\qquad \dashrightarrow \sf a_n = a + \bigg\lgroup n - 1 \bigg\rgroup d \\\\\\ \qquad \dashrightarrow \sf a_5 = 24 + \bigg\lgroup 5 - 1 \bigg\rgroup (-4) \\\\\\ \qquad \dashrightarrow \sf a_5 = 24 + \bigg\lgroup 4 \bigg\rgroup (-4) \\\\\\ \qquad \dashrightarrow \sf a_5 = 24 - 16 \\\\\\ \qquad \dashrightarrow \sf \underline {\boxed{\pmb{\frak{ a_5 \: ( \: or \: 5^{th} \: Term \:)\:=\; 8 \:}}}}\:\:\bigstar$

$\qquad \therefore \:\underline {\sf Hence ,\: The \: 5^{th} \: Term \: of \: an \: A.P \: is \:\pmb{\bf 8\:}\;.}$