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 sn, the sum of first n terms of an AP is given by sn= (3n2-4n), 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 S1 , S2 , S3 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 S1+S3= 2S2.
which term of an AP 3,15,27,39............ will 132 more than its 54th 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 2n2 + 5n, then find its nth 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 sn, the sum of first n terms of an AP is given by sn= (3n2-4n), 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 S1 , S2 , S3 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 S1+S3= 2S2.
which term of an AP 3,15,27,39............ will 132 more than its 54th 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 2n2 + 5n, then find its nth 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).