Question

If the sum of first 16 terms of an AP is 760 and
its first term is 10, find the 30th term.​

Answer 1

Answer:

let the common difference is b

here the 1st term is a=10

sum of the n terms is n/2{2a+(n-1).b}

sum of 16 terms is

16/2{2.10+(16-1).b=760

=> 8{20+15.b}=760

=> 20+15b=760/8=95

=> 15b=95-20=75

=> b=75/15=5

so the 30th term is a+(30-1).b

=10+29.5=10+145=155

Answer 2

Answer:

= 155

Step-by-step explanation:

given: first term = 10

Sum of 16 terms = 760

therefore S16 = n/2 [2(a) + (n-1)d]

760= 16/2 [2(10)+(16-1)d]

760= 8 [20+15d]

760/8 = 20+15d

95= 20 +15d

15d= 95-20

d= 75/15

d= 5

according to question

30th term

T30 = a + (n-1)d

T30 = 10 + (30-1) 5

T30 = 10 + (29)5

T30 = 10 + 145

T30 = 155

answer.