The ratio of the sum of n terms of two A.P.s is (7n+1):(4n+27).Find the ratio of their mth term.

1. The ratio of the sums of m and n terms of an AP is m² : n^2. show that the ratio of the m^th and n^th terms is (2m-1) : (2n-1)
2. In an AP, the sum of first n terms is 3n²/2 + 5n/2, find its 25^th term.

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. Find the numbers.

If the m^th term of an A.P. be 1/n and n^th term be 1/m, then show that its (mn)^th term is 1.

IF THE SUM OF FIRST m TERMS OF AN AP IS n. AND THE SUM OF FIRST n TERMS IS m, THEN SHOW THAT THE SUM OF ITS FIRST( m+n) TERMS IS -(m+n).

Maths Project :make designs using arithmetic  progression
 

if the pth term of an A.P is q and the qth term is p, prove that its nth term is (p+q-n)

the sum of the first n terms of an AP is given by s_n=3n²-n , determine the AP and its 25^thterm.

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What is the explaination for the double answer?

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If pth , qth and rth term of an AP are a,b,c respectively , then show that

(a-b)r +(b-c)p + (c-a)q = 0

Please help me with this question asap .

 150 workers were engaged to finish a piece of work in a certain no. of days. 4 workers droped d 2nd day, 4 more workers droped the 3rd day n so on..... it takes 8 more days to finish the work now ..find d no. of days in vch the work waz completed....

An AP consists of 37 terms. The sum of the three middle most terms is 225 and the

find the 6th term from the end of the AP 5,2, -1 , -4................. -31 .

If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289.Find the sum of first n terms

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plzz reply with solution

How many 3 digit numbers are such that when divided by 7, leave a remainder 3 in each case.

find x so that 2x+1, x²+x+1&3x²- 3x+ 3 are the consecutive terms of an A.P.

the ratio of the sum of n terms of two AP's is (7n+1):(4n+27).find the ratio of their m th terms.

Answer: let a1 , a2 be the 1st terms and d1 , d2 the common differences of the two given A.P's. then the sums of their n terms are given by

Sn = n/2 {2.a1+(n-1)d1} and Sn' = n/2{2. a2 +(n-1)d2}

Sn/Sn' = n/2{2.a1+(n-1)d1} / n/2{2.a2 + (n-1)d2}

Sn / Sn' = 2.a1+(n-1)d1 / 2.a2 + (n-1)d2

it is given that

Sn / Sn' = 7n+1 / 4n + 27

2.a1 + (n-1)d1 / 2.a2 + (n-1)d2 = 7n+1 / 4n+27 ........................ (i)

To find the ratio of the mth terms of the two given AP's , we replace n by (2m-1) in equation (i)

therefore, 2.a1 + (n-1)d1 / 2.a2 + (n-1)d2 = 7(2m-1) + 1 / 4(2m-1) + 27

a1 + (m-1)d1 / a2 + (m-1)d2 = 14m - 6 / 8m + 23

Hence, the ratio of the mth terms of the two A.P's is (14m - 6) : (8m + 23)

My question is, why it has been assumed (2m-1) in the place of 'n' ? has it been arrived from solving with the help of a formula or is it a mere assumption? if it is simply an assumption, why it should be assumed as (2m-1) ? why not 2m or (m - 1)

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urgent

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Hint:

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