if tn denotes the nth term of the series 2+3+6+11+18+...........,then find t50.
if it is an AP,Sn=Pn2 and Sm =Pm2 m is not equal to n in an A.P, where Srdenotes the sum of r terms of A.P. find Sp.
let Sn denote the sum of first n terms of an A.P. if S2n=S3n.then find S3nS2n.
find the 10th term common between the series 3+7+11+..... and 1+6+11+....
if Sr denotes the sum of first r terms of an A.P.,then find S3r-Sr-1/S2r-S2r-1.
show that the sum of an A.P whose first term is'a',the second tern is 'b' and the last term is'c', is equal to (a+c)(b+2c-2a)/2(b-a).
sum of 1st n terms of 3 A.p. s' are S1 S2 S3. if the first term of each if the three A.P.s is 1 and common difference are 1,2,3 respectively, then prove that S1 + S3= 2S2.
in an A.P, the pth term and qth term and rth term are p, q, r respectively. show that p(q-r) + q(r-p) +r (p-q) =0.
in an A.P xth term is 1/y and y th term is 1/x. find the sum of the first xy terms.
find the sum to n terms of the series (1/1+12+14)+(2/1+22 + 24) + 3/1+32+34................
if a1,a2,a3,.....,an be in AP of non zero terms, then prove that 1/a1a2 +1a2a3........+1/an- 1an=n-1/a1an.
if a,b,c are in A.p be 1/n and its nth term be 1/m, then show that its
(mn)th term is 1.
the ratio of the sum of n terms of two A.P 's is (2n+1)(5n+3).find the ratio of their 10th terms.
An A.P has 2n+1 terms. Let S0 represent the sum of its odd terms and Se represent the sum of its even terms. prove that S0 - Se =A + nd ,where A iis the first term d is the common difference.
Prove that Sn,S2n- sn, S3n-S2n, S4n-S3n are in A.P find out the common differnce.
A positive interger has three digits which are in A.P and their sum is 15. the number obtained on reversing the digits is 594 less than the original numbver. find the number
the ratio of sum of m terms and n terms of an A.P be m2 n2.prove that the ration of mth and nth terms will be
(2m-1) (2n-1).
if the pth term of an A.P is a and qth term is b, show that the sum of (p+q) terms is p+q/2[a+b+a-b/p-q].
Answers
Answer 1:
Now, This difference of the terms of this series is in A.P.
3 - 2 = 1
6 - 3 = 3
11 - 6 = 5
18 - 11 = 7
So, the series obtained from the difference = 1,3,5,7,...
and to get back the original series we need to add the difference back to 2.
2+1 = 3,
2+1+3 = 6,
2+1+3+5= 11,
2+1+3+5+7 = 18 and so on.
So, we can say that nth term of our given series ( 2 + 3 + 6 + 11 + 18+.... ) is = Sum of ( n - 1 ) term of series ( 1,3,5,7,... ) + 2
So, we need to calculate the sum of 49 terms of the series 1,3,5,7,9,11,..
As we know formula for nth term in A.P.
Sn = n/2[ 2a + ( n - 1 ) d ]
Here a = first term = 1 , n = number of term = 49 and d = common difference = 2 , So
Sn = 49/2[ 2( 1 ) + ( 49 - 1 ) 2 ] = 49 [ 1 + ( 49 - 1 ) ] = 492
Hence, Sum of 49 terms of series 1,3,5,7,9,11,.. = 492
Now, to get the T50 term.. add 2+ sum of the 1+3+5+7+..+97
So ,
T50 of series 2 + 3 + 6 + 11 + 18+....... = 2 + 492 = 2 + 2401 = 2403
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