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JEE Advanced · Mathematics · 5. Sequences & Series

Let a1,a2,a3, be an arithmetic progression with a1=7 and common difference 8. Let T1,T2,T3, be such that T1=3 and Tn+1-Tn=an fo n1. Then, which of the following is/are TRUE?

  1. A T20=1604
  2. B k=120Tk=10510
  3. C T30=3454
  4. D k=130Tk=35610
Verified Solution

Answer & Solution

Correct Answer

(B) k=120Tk=10510

Step-by-step Solution

Detailed explanation

Given,
an=7+n-18 and T1=3
Also Tn+1=Tn+an
Now Tn=Tn-1+an-1
Putting n=1 we get, 
T2=T1+a1
Now putting n=2 we get,
T3=T2+a2T3=T1+a1+a2
And so on we get,
 Tn+1=T1+a1+a2++an
 Tn+1=T1+n227+n-18
 Tn+1=T1+n4n+3     1
Now for n=19 we get,
T20=3+1979=1504
And for n=29 we get,
T30=3+29119=3454 {option C is correct}
Now finding sum  we get,
\(\sum_{k=1}^{20} T_k=3+\sum_{k=2}^{20} T_k=3+\sum_{k=1}^{19}(3~+\) \(4 n^2+3 n)\)
=3+319+319202+41920396
=3+10507=10510 {option B is correct}
And Similarly k=130Tk=3+k=1294n2+3n+3=35615
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