Question

35. The sum of first n terms of two APs are in the ratio (3n+8):(7n +15).
Find the ratio of their 12th terms.​

Answer 1

Answer:

4/9

Step-by-step explanation

Let nth term of first AP = 3n+8…(i)

First term, a = 3+8 = 11 (put n = 1 in (i))

Second term = 6+8 = 14

Common difference, d = 14-11 = 3

12th term = a+11d

= 11+11(3)

= 44

Let nth term of second AP = 7n+15 ..(ii)

First term, a = 7+15 = 22 (put n = 1 in (ii))

Second term = 14+15 = 29

Common difference, d = 29-22 = 7

12th term = a+11d

= 22+11(7)

= 22+77

= 99

The ratio of their 12th terms = 44/99 = 4/9

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Answer 2

Answer:

7:16

PLZ:

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Step-by-step explanation:

(3n+8):(7n +15)=n/2(11+3(n-1)):n/2(22+7(n-1))

                         by comparing with Sn=n/2[2a+(n-1)d]

we get

1st AP      2nd AP

a=11/2       a=11

d=3           d=7

ratio of their 12th terms

(11/2+(12-1)3):(11+(12-1)7)

(11/2+33):(11+77)

(77/2):(88)

7:16