if 5th term of ap is 13 and sum of 15 terms is 285 find 14th term or n term
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5th term = 13
=> a+(5-1)d = 13
=> a+4d = 13. ____(1)
sum of 15 terms = 285
=> S15 = 285
=> 15/2 { 2a +(15-1) d} = 285
=> 2a+14d = 285×2 /15
=> 2(a+7d) = 38
=> a+7d = 38/2
=> a+7d = 19. _____(2)
eq(1) - eq (2)
we get :
-3d = -6
=> d = 2
put this value in equation (1), we get:
a= 5
a14 = a+(n-1)d
= 5+(14-1)2
= 5+26
= 31
and,
nth term = a+(n-1) d
= 5+(n-1)2
= 5+2n -2
= 3+2n
=> a+(5-1)d = 13
=> a+4d = 13. ____(1)
sum of 15 terms = 285
=> S15 = 285
=> 15/2 { 2a +(15-1) d} = 285
=> 2a+14d = 285×2 /15
=> 2(a+7d) = 38
=> a+7d = 38/2
=> a+7d = 19. _____(2)
eq(1) - eq (2)
we get :
-3d = -6
=> d = 2
put this value in equation (1), we get:
a= 5
a14 = a+(n-1)d
= 5+(14-1)2
= 5+26
= 31
and,
nth term = a+(n-1) d
= 5+(n-1)2
= 5+2n -2
= 3+2n
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