Question

The fifth term of the AP is 13 and its 9th term is 21. Find the AP and the sum of first 10 terms.​

Answer 1

GIVEN :–

• 5th term of A.P. is 13.

• 9th term of A.P. is 21.

TO FIND :–

• A.P. = ?

• Sum of first terms = ?

SOLUTION :–

• We know that nth term of A.P. is –

$\\ \bf \implies \large{ \boxed{ \bf T_n = a+(n-1)d}}$

• According to the first condition –

$\\ \bf \implies T_5 =13$

$\\ \bf \implies a + (5 - 1)d =13$

$\\ \bf \implies a +4d =13$

$\\ \bf \implies a=13 - 4d \: \: \: \: - - - eq.(1)$

• According to second condition –

$\\ \bf \implies T_9 =21$

$\\ \bf \implies a + (9- 1)d =21$

$\\ \bf \implies a+8d=21$

• Using eq.(1) –

$\\ \bf \implies 13 - 4d+8d =21$

$\\ \bf \implies 13 + 4d=21$

$\\ \bf \implies 4d=21 - 13$

$\\ \bf \implies 4d=8$

$\\ \bf \implies d=\cancel\dfrac{8}{4}$

$\\ \bf \implies \large{ \boxed{ \bf d=2}}$

• Again Using eq.(1) –

$\\ \bf \implies a=13 - 4(2)$

$\\ \bf \implies a=13 -8$

$\\ \bf \implies \large{ \boxed{ \bf a = 5}}$

• A.P. $\bf \:\implies 5,7,9,11........$

• We also know that –

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

• Put the values –

$\\ \bf \implies S_{10} = \dfrac{10}{2}[2(5)+(10 - 1)(2)]$

$\\ \bf \implies S_{10} = (5)[10+(9)(2)]$

$\\ \bf \implies S_{10} = (5)[10+18]$

$\\ \bf \implies S_{10} = (5)[28]$

$\\ \bf \implies \large{ \boxed{ \bf S_{10} =140}}$

Answer 2

$\large\bf{\underline{\underline\red{Given:-}}}$

• The fifth term of the AP is 13

• The ninth term of AP is 21.

$\large\bf{\underline{\underline\blue{To\:Find:-}}}$

• The AP

• The sum of first 10 terms of the AP.

$\large\bf{\underline{\underline\green{Solution:-}}}$

As we know that :-

Sum of n terms is given by the formula.

$\boxed{ \rm \leadsto a_{n} = a + (n - 1)d }$

Here:-

• n = number of terms

• a = first term

• d = common difference.

The fifth term

$\rm \implies a_{5} = 13$

$\rm \implies a + (5- 1)d = 13$

$\rm \implies a + 4d = 13......(i)$

The ninth term

$\rm \implies a_{9} = 21$

$\rm \implies a + (9- 1)d = 21$

$\rm \implies a + 8d = 21......(ii)$

Subtracting equation (i) from (ii)

$\rm \implies a + 8d - (a + 4d) = 21 - 13$

$\rm \implies a + 8d - a - 4d= 8$

$\rm \implies 4d= 8$

$\rm \implies d= 2$

Substituting d = 2 in equation (i)

$\rm \implies a + 4d= 13$

$\rm \implies a + 4(2)= 13$

$\rm \implies a +8= 13$

$\rm \implies a = 5$

We have :- a = 5 and d = 2

We need to find the sum if 10 terms:-

Sum of n terms is given by the formula:-

$\boxed{\rm \leadsto S_{n} = \frac{n}{2} \bigg \{2a + (n - 1)d \bigg \} }$

Here:-

• n = 10

• a = 5

• d = 2

${\rm \leadsto S_{10} = \dfrac{10}{2} \bigg \{2 \times 5+ (10- 1)2 \bigg \} }$

${\rm \implies S_{10} = 5 \big \{ 10+ 18 \} }$

$\rm \implies S_{10} = 5 \times 28$

$\rm \implies S_{10} = 140$

$\large \underline{ \boxed{ \rm \therefore Sum \: of \: 10 \: terms= 140}}$