Question

The sums of n terms of two arithmetic progressions are in the ratio 5 n + 4: 9 n + 6. Find the ratio of their 18 th terms.

Answer 1

The ratio of the sum of two arithmetic progressions is

5n+4 9n+6 .

Let a 1 , a 2 and d 1 , d 2 be the first terms and common differences of the first and second arithmetic progressions respectively.

The formula for the sum of n terms in an A.P. is given by,

S n = n 2 [ 2a+( n−1 )d ](1)

Let S 1 , S 2 be the sum of the first and second arithmetic progression.

Substitute the values to obtain the sum of first arithmetic progression.

S 1 = n 2 [ 2 a 1 +( n−1 ) d 1 ]

Similarly, substitute the values to obtain the sum of second arithmetic progression.

S 2 = n 2 [ 2 a 2 +( n−1 ) d 2 ]

As per the question,

S 1 S 2 = 5n+4 9n+6 (2)

Substitute the values of S 1 and S 2 in equation (2).

n 2 [ 2 a 1 +( n−1 ) d 1 ] n 2 [ 2 a 2 +( n−1 ) d 2 ] = 5n+4 9n+6 2 a 1 +( n−1 ) d 1 2 a 2 +( n−1 ) d 2 = 5n+4 9n+6

Substitute the value of n=35 in the above expression.

2 a 1 +( 35−1 ) d 1 2 a 2 +( 35−1 ) d 2 = 5×35+4 9×35+6 2 a 1 +34 d 1 2 a 2 +34 d 2 = 175+4 315+6 2 a 1 +2×17 d 1 2 a 2 +2×17 d 2 = 179 321 2×( a 1 +17 d 1 ) 2×( a 2 +17 d 2 ) = 179 321

Further simplify the above expression.

( a 1 +17 d 1 ) ( a 2 +17 d 2 ) = 179 321 (3)

Let T 1 and T 2 be the 18 th term of first and second A.P.

The formula to find the terms in an A.P. is given by,

T n =a+( n−1 )d

Substitute the value of n as 18 in the above expression to obtain the 18 th term of first and second A.P.

T 1 = a 1 +( 18−1 ) d 1 = a 1 +17 d 1 T 2 = a 2 +( 18−1 ) d 2 = a 2 +17 d 2

The ratio of the 18 th term of first and second A.P. is given as,

T 1 T 2 = a 1 +17 d 1 a 2 +17 d 2

Substitute the values from equation (3).

T 1 T 2 = 179 321

Thus, the ratio of the 18 th term of first and second A.P. is 179 321 .