Advertisements
Advertisements
प्रश्न
If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
Advertisements
उत्तर
Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P’s.
Then we have `S_n=n/2[2a_1+(n-1)d_1] " and " S_n=n/2[2a_2+(n-1)d_2]`
∴ `S_n/S_n=(n/2[2a_1+(n-1)d_1])/(n/2[2a_2+(n-1)d_2])=(2a_1+(n-1)d_1) /(2a_2+(n-1)d_2)`
It is given that `S_n/S_n=(7n+1)/(4n+27)`
`:.(2a_1+(n-1)d_1)/(2a_2+(n-1)d_2)=(7n+1)/(4n+27) "....(1)"`
To find the ratio of the mth terms of the two given A.P.'s replace n by (2m-1) in equation (1).
`:.(2a_1+(2m-1-1)d_1)/(2a_2+(2m-1-1)d_2)=(7(2m-1)+1)/(4(2m-1)+27)`
`:.(2a_1+(2m-1)d_1)/(2a_2+(2m-2)d_2)=(14m-7+1)/(8m-4+27)`
`:.(a_1+(m-1)d_1)/(a_2+(m-1)d_2)=(14m-6)/(8m+23)`
Hence, the ratio of the mth terms of the two A.P's is 14m-6 : 8m + 23
APPEARS IN
संबंधित प्रश्न
The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)
In an AP Given a12 = 37, d = 3, find a and S12.
Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………
The sum of the first n terms in an AP is `( (3"n"^2)/2 +(5"n")/2)`. Find the nth term and the 25th term.
The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.
Find the first term and common difference for the A.P.
0.6, 0.9, 1.2,1.5,...
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.
How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?
Find the sum of first seven numbers which are multiples of 2 as well as of 9.
