The sum of n terms of two ap are in ratio (5n+4) : (9n+6) Find the ratio of their 18th term?

plzz reply with solution

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)

If s_{n}, the sum of first n terms of an AP is given by s_{n}= (3n^{2}-4n), then find its nth term.

If the sum of m terms of an A.P is the same as the sum of n terms of the same A.P, show that the sum of its (m+n) terms is zero.

what are the uses of arithmetic progressions in daily life? i want atmost 10, pls help me as soon as possible

if the nth term of an AP is (2n+1) find the sum of first n terms of the AP

sum of first p , q and r terms of an A.P. are a , b and c respectively. Prove that :

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

If S_{1} , S_{2} , S_{3} are the sum of n terms of three AP's the first term of each being unity and respective common diffrence being 1 , 2 , 3_ _ _ _ _. Prove that S_{1}+S_{3}= 2S2.

which term of an AP 3,15,27,39............ will 132 more than its 54^{th} term

Divide 56 into 4 parts which are in AP such that hte ratio of product of extremes to the product of mean is 5 : 6? Hoping for a quick response......plz.....[ May be within Sunday ]

if sum of n terms of an AP 2n^{2} + 5n, then find its n^{th} term.

In an AP,the sum of first n term is 3n2/2+13n/2

find its 25th term

solve the solution 1+4+7+10.............+x = 287

if Sn denotes the sum of first n terms of an A.P., prove that S12=3(S8-S4).

The sum of n terms of two ap are in ratio (5n+4) : (9n+6) Find the ratio of their 18th term?

plzz reply with solution

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)

If s

_{n}, the sum of first n terms of an AP is given by s_{n}= (3n^{2}-4n), then find its nth term.If the sum of m terms of an A.P is the same as the sum of n terms of the same A.P, show that the sum of its (m+n) terms is zero.

what are the uses of arithmetic progressions in daily life? i want atmost 10, pls help me as soon as possible

if the nth term of an AP is (2n+1) find the sum of first n terms of the AP

sum of first p , q and r terms of an A.P. are a , b and c respectively. Prove that :

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

If S

_{1}, S_{2}, S_{3}are the sum of n terms of three AP's the first term of each being unity and respective common diffrence being 1 , 2 , 3_ _ _ _ _. Prove that S_{1}+S_{3}= 2S2.which term of an AP 3,15,27,39............ will 132 more than its 54

^{th}termDivide 56 into 4 parts which are in AP such that hte ratio of product of extremes to the product of mean is 5 : 6? Hoping for a quick response......plz.....[ May be within Sunday ]

if sum of n terms of an AP 2n

^{2}+ 5n, then find its n^{th}term.In an AP,the sum of first n term is 3n2/2+13n/2

find its 25th term

solve the solution 1+4+7+10.............+x = 287

if Sn denotes the sum of first n terms of an A.P., prove that S12=3(S8-S4).