Question

find the ratio of 15th term of AP 3n+4 to 9 term of ap.p+3 ​

Answer 1

Answer:

ANSWER

For 1

st

AP

Let first term be a common difference be d

sum of n terms s

n

=

2

n

(2a+(n−1)d)

n

th

term a

n

=a(n−1)d

Similarly for 2

nd

AP

Let first term =A common difference be D

S

n

=

2

n

(2A+(n−1)D) & n

th

term =A

n

=A+(n−1)D

We need ratio of 12

th

term

i.e.,

A

12

ofsecondAP

a

12

offirstAP

=

A+(12−1)D

a+(12−1)d

=

a+11D

a+11d

It is given that

Sumofntermsof2

nd

AP

Sumofntermsof1

st

AP

=

7n+15

3n+8

∴

2A+(n−1)D

2a+(n−1)d

=

7n+15

3n+8

………….(1)

2(A+(

2

n−1

)D)

2(a+(

2

n−1

)d)

=

7n+15

3n+8

………………(1)

we need to find

A+11D

a+11d

Hence

2

n−1

=11

n−1=22

n=23

Putting n=23 in (1)

A+(

2

23−1

)D

a+(

2

23−1

)d

=

7×23+15

3×23+18

∴

A+11D

a+11d

=

16

7

Hence ratio of their 12

th

term is

16

7

i.e., 7:16

Answer 2

Answer:

$\frac{7}{16}$

=7:16 ANSWER

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