Question

Find the sum of 15 terms of an A. P. whose fifth and ninth terms are 26 and 42 respectively.​

Answer 1

Given:

• Number of terms = 15
• Fifth term ($a_5$) = 26
• Ninth term ($a_9$) = 42

Formulas to be used:

$\fbox{\sf{General\: Term \:of \:AP \:= a+(n-1)d}}$

$\sf{Sum \: of \: n \: terms \: of \: AP \:=\dfrac{n}{2}(2a+(n-1)d)}$

Solution:

Fifth Term = a+(5-1)d

$\large{⇒}$a+4d = 26 ...(1)

Ninth Term = a+(9-1)d

$\large{⇒}$ a+8d = 42 ... (2)

Subtracting equation (1) from (2), we get:

$\large{⇒}$4d = 16

$\large{⇒}$d = 4

Substituting the value of d in equation (1),

$\large{⇒}$a + 16 = 26

$\large{⇒}$a = 26-16

$\large{⇒}$a = 10

Substituting the value of a and d in sum formula,

Sum = $\dfrac{15}{2}(2\times10+(15-1)4)$

$\large{⇒}$$\dfrac{15}{2}(20+56)$

$\large{⇒}$$\dfrac{15\times76}{2}$

$\large{⇒}$$15\times38$

$\therefore$Sum of given AP = 570