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ICSE Class 10 Answered

13 th ter of an AP is 4 times its 3rd term .if 5 th term is 16 then find sum of its first ten terms   the sum of first 7 term of Ap is 182 if its 4 th and 17 th term are in ratio1:5,find AP
Asked by chahaljain03 | 30 Dec, 2018, 21:49: PM
answered-by-expert Expert Answer
Let a be the first term and d be the common difference of given A.P
 
13th term t13 = a+12d 
3rd term t3 = a+2d
we are given : a+12d = 4(a+2d)  or 3a = 4d  or a/d = 4/3    ...........................(1)
 
5th term, t5 = a+4d = 16   ...........................(2)
 
Using eqn.(1) in eqn.(2)    (4/3)d+4d = 16  .....(3)
 
solving eqn.(3) for d, we get d = 3 ; hence a =(4/3)×3 = 4
 
Hence sum of  first ten terms, S10 = (10/2)[ 2×4 + 9×3 ] = 175
 
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Second question:
 
Sum of seven terms = S7 = (7/2) [ 2a + 6d ] = 182   or a+3d = 26 .....................(1)
 
begin mathsize 12px style t subscript 4 over t subscript 17 space equals space fraction numerator a space plus space 3 d over denominator a plus 16 d end fraction space equals space 1 fifth space space space o r space space space 5 a space plus space 15 d space equals space a space plus space 16 d space space space o r space space space space a space equals space d over 4 space end style
By substituting a in eqn.(1), we get a =2 and d=8
 
Hence A.P is with first term a =2 and common difference d = 8
Answered by Thiyagarajan K | 31 Dec, 2018, 11:00: AM

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