Question

The sums of n terms of two arithmetic progressions are in the ratio of

(7n + 1) : (4n + 27). Find the ratio of their 11th terms​

Answer 1

Answer:

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Step-by-step explanation:

Let the two series' be T

n

and T

n

′

with first terms a and a

′

and common differences d and d

′

The ratio of the sums of the series' S

n

and S

n

′

is given as,

S

n

′

S

n

=

[n/2][2a

′

+[n−1]d

′

]

[n/2][2a+[n−1]d]

=

4n+27

7n+1

Or,

a

′

+[(n−1)/2]d

′

a+[(n−1)/2]d

=

4n+27

7n+1

...(1)

We have to find,

T

11

′

T

11

=

a

′

+10d

′

a+10d

Choosing (n−1)/2=10 or n=21 in (1) we get

T

11

′

T

11

=

a

′

+10d

′

a+10d

=

4(21)+27

7(21)+1

=

111

148

=

3

4

.