Maths-
General
Easy

Question

Let s subscript 1 comma s subscript 2 comma s subscript 3 horizontal ellipsis horizontal ellipsis and t subscript 1 end subscript comma t subscript 2 end subscript comma t subscript 3 end subscript horizontal ellipsis horizontal ellipsis are two arithmetic sequences such that s subscript 1 end subscript equals t subscript 1 end subscript not equal to 0 semicolon s subscript 2 end subscript equals 2 t subscript 2 end subscript and stretchy sum from i equals 1 to 10 of   s subscript i end subscript equals stretchy sum from i equals 1 to 15 of   t subscript i end subscript then the value of fraction numerator s subscript 2 end subscript minus s subscript 1 end subscript over denominator t subscript 2 end subscript minus t subscript 1 end subscript end fraction is

  1. 8/3    
  2. 3/2    
  3. 19/8    
  4. 2    

hintHint:

use the formula Sn = (n/2)( 2 a1 + (n-1)d)
for the summation.

The correct answer is: 19/8


    19/8
    given, s1=t1 ; s2 = 2t2
    let s2- s1 = d1 and t2-t1 = d2
    s2-s1 = 2t2-t1 = d1
    given,
    summation (i=1 to 10) si = summation ( i= 1 to 15) ti
    => (10/2)( 2s1 + 9 d1) = (15/2) (2t1 + 14 d2)
    on solving, we get
    2s1 +9d1 = 3t1 + 21 d2
    substituting the values of s1, d1 and d2 into this equation, we get
    2t1 + 9(2t2-t1)=3t1+21(t2-t1)
    or
    3t2=11t1
    substituting the  value of t2 into the equation (s2-s1)/(t2-t1) we get
    (2t2-t1)/(t2-t1)= (22/3-1)/(11/3-1)
    = 19/8

    an A.P is a sequence of mathematical terms which have a common difference with their adjacent elements

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