Question

32. How many terms of an A.P. 24, 20,16.... must be taken so that the sum be 72?
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Answer 1

Answer:

Here, It is stated that an A.P. 24, 20,16.... is added up to 72. So, here it has asked us to find the number of terms must be taken to add up to 72.

For that, first we will find out the common difference, there after we are going to use the sum of nth term formula in order to get the value of n.

So, let's solve it ::

Formula Used :

$\bullet \: \: {\boxed{ \bf{ S_n = \dfrac{n}{2}[2a + (n - 1) \times d] }}}$

Answer :

• First Term → 24

For Common Difference → $\sf A_2 - A_1$

Here,

$\sf A_2 = Second \: Term$

$\sf A_1 = First \: Term$

Now,

Applying Formula ::

$\dashrightarrow \: \: \: \sf S_n = \dfrac{n}{2}[2a + (n - 1) \times d]$

From Given ;

$\sf a = 24$

$\sf d = - 4$

$\sf S_n = 72$

Substituting the respective values, we get ;;

$\dashrightarrow \: \: \: \sf 72 = \dfrac{n}{2}[2(24) + (n - 1) \times (-4)]$

$\dashrightarrow \: \: \: \sf 72 = \dfrac{n}{2} [48+ (n - 1) \times (-4)]$

$\dashrightarrow \: \: \: \sf 72 = \dfrac{n}{2} [48 - (n - 1) \times (4)]$

$\dashrightarrow \: \: \: \sf 72 = n[24 - (n - 1)2]$

$\dashrightarrow \: \: \: \sf 72 = [24n - n(n - 1)2]$

$\dashrightarrow \: \: \: \sf 72 = [ 24n - (n^{2} + n)2]$

$\dashrightarrow \: \: \: \sf 72 = [24n - 2n^{2} + 2n]$

$\dashrightarrow \: \: \: \sf 72 = 26n - 2n^{2}$

Transposing the terms on L. H. S. side ::

$\dashrightarrow \: \: \: \sf - 26n + 2n^{2} + 72 = 0$

Arranging in General Form :

$\dashrightarrow \: \: \: {\underline{\bf ax^{2} + bx + c = 0}}$

$\dashrightarrow \: \: \: \sf 2n^{2} - 26n + 72 = 0$

Dividing whole equation By 2 , to make it easier to split.

$\dashrightarrow \: \: \: \sf [2n^{2} - 26n + 72 = 0] \div 2$

$\dashrightarrow \: \: \: \sf n^{2} - 13n + 36 = 0$

By Using Middle - Term Splitting method :

$\dashrightarrow \: \: \: \sf n^{2} - (4 +9)n + 36 = 0$

$\dashrightarrow \: \: \: \sf n^{2} - 4n - 9n + 36 = 0$

Taking common :

$\dashrightarrow \: \: \: \sf n(n- 4) - 9(n - 4) = 0$

$\dashrightarrow \: \: \: \sf (n- 4) (n - 9) = 0$

$\dashrightarrow \: \: \: \sf (n- 4) = 0$

$\dashrightarrow \: \: \: \bf n = 4$

$\dashrightarrow \: \: \: \sf (n - 9) = 0$

$\dashrightarrow \: \: \: \bf n = 9$

${\boxed{\therefore {\sf {Number \: of \: Terms = {\red{4 \: Or \: 9}} }}}}$

Hence, The no. of terms must be taken to give sum of 72 → 4 Or 9.

Answer 2

Given -

• A.P. → 24, 20, 16 ....

⠀

To find -

• Number of terms that must be taken for the A.P. so that the sum be 72.

⠀

Solution -

We have a A.P. with first term (a) = 24 and common different (d) = -4.

Here it is given that sum should be 72, therefore

• $\bf{S_n = 72}$

⠀

We need to find the number of terms (n). For finding it, we will use the given formula.

⠀

$\underline{\boxed{\red{S_n = \dfrac{n}{2} [2a + (n - 1)d]}}}$

⠀

★ Putting the all values

$\tt\dashrightarrow{72 = \dfrac{n}{2} [ (2 \times 24) + (n - 1)(-4)]}$

⠀

$\tt\dashrightarrow{144 = n[48 - 4n + 4]}$

⠀

$\tt\dashrightarrow{144 = n[52 - 4n]}$

⠀

$\tt\dashrightarrow{144 = 52n - 4n^2}$

⠀

$\tt\dashrightarrow{4n^2 - 52n + 144 = 0}$

⠀

$\tt\dashrightarrow{n^2 - 13n + 36 = 0}$

⠀

Now, we have a quadratic equation, we will split the middle term for finding the value of n.

$\tt\dashrightarrow{n^2 - 9n - 4n + 36 = 0}$

⠀

$\tt\dashrightarrow{n(n - 9) - 4(n - 9) = 0}$

⠀

$\tt\dashrightarrow{(n - 9) (n - 4) = 0}$

⠀

We get,

• n = 9
• n = 4

⠀

Hence,

• Number of terms should be 9 or 4.

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