Question

The sum of six terns of an arithmetic progression is 9,the sum of 9 terms is 6.The sum of fifteen term is

Answer 1

It is given that the sum of six terms of an arithmetic progression is 9.

That is $S_6= 9$

Sum of 'n' terms of an arithmetic progression = $\frac{n}{2}[2a+(n-1)d]$

Since, $S_6= 9$

$\frac{6}{2}[2a+(6-1)d] = 9$

$3[2a+5d] = 9$

$2a+5d = 3$  

$a = \frac{3-5d}{2}$   (Equation 1)

Since, $S_9= 6$

$\frac{9}{2}[2a+(9-1)d] = 6$

$9[2a+8d] = 12$

$2a+8d = \frac{12}{9}$

$2a+8d = \frac{4}{3}$

$6a+24d = 4$

$3a+12d = 2$ (Equation 2)

Substituting the value of 'a' from equation 1, we get

$3 [\frac{3-5d}{2}] + 12d = 2$

$\frac{9-15d}{2} + 12d = 2$

$\frac{{9-15d}+ 24d}{2} = 2$

$\frac{9+9d}{2} = 2$

${9+9d} = 4$

$9d = -5$

$d = \frac{-5}{9}$

Since, $a = \frac{3-5d}{2}$

$a = \frac{3-5(\frac{-5}{9})}{2}$

$a = \frac{26}{9}$

Now, we have to determine the sum of fifteen terms,

$S_{15} = \frac{15}{2}[2(\frac{26}{9}) + 14(\frac{-5}{9})]$

$S_{15} = \frac{15}{2}[\frac{-18}{9}]$

$S_{15} = -15$

So, the sum of fifteen term of an AP is -15.