Question

an AP S15=360 and S14=305 then the a15 is​

Answer 1

Answer:

Hi there, here we have,

$S_{15} = 360\\and\\S_{14}= 305$

So,

We know that formula for sum of an AP is

S = n/2(2a+(n-1)d)

where n = No. of terms

          a = The first Term

          d = Common difference

so in  $S_{15}$ we have n = 15

and in $S_{14}$ we have n = 14

so simplifying we have

$15/2 (2a +(15-1)d) = 360\\=> 15(2a+14d) = 720\\=> 2a +14d = 720/15\\=> a+7d = 360/15\\=> a +7d = 24$(1)

and

$14/2(2a+(14-1)d) = 305\\=> 7(2a+13d) = 305\\=>14a+91d = 305$(2)

So simplifying (1) and (2),

We have,

$a+7d = 24\\14a+91d = 305$

and by elimination we can solve for a and d

$13(a+7d=24)\\14a+91d=305\\\\13a+91d = 312\\14a +91d=305\\\\-1(13a+91d=312)\\14a+91d = 305\\\\-13a-91d = -312\\14a+91d = 305\\\\solving,\\a=-7\\\\so,\\\\(-7) + 7d = 24\\7d = 31\\d = 31/7$

so a = -7

d = 31/7

so a15 is

$-7+(15-1)*31/7 \\=> -7+2*31\\=>-7+62\\=> 55$

a15 is 55

Hope it helps