Question 10 - Jee advanced Math 2022 P1 Questions with Solutions

Let \(a_{1}, a_{2}, a_{3}, ….\) be an arithmetic progression with \(a_{1}=7\) and common difference \(8\). Let \( T_{1}, T_{2}, T_{3},…\) be such that \(T_{1} = 3\) and \(T_{n+1} - T_{n} = a_{n}\) for \(n \geq 1\). Then, which of the following is/are TRUE?

A) \(T_{20} =1604\)

B) \(\sum_{k=1}^{20} T_{k} = 10510\)

C) \(T_{30} = 3454\)

D)  \(\sum_{k=1}^{30} T_{k} = 35610\)

Sol : 

\(a_{1}, a_{2}, a_{3}, ….\) is an arithmetic progression.

\(\implies a_{n} = a_{1} + (n - 1)d\)

Also,

\(T_{n+1} - T_{n} = a_{n}\)

\(\implies T_{n+1} = T_{n} + a_{n}\) 

i.e., \(T_{2} = T_{1} + a_{1}; \:  T_{3} = T_{2} + a_{2}; \: T_{4} = T_{3} + a_{3}; \:\) and so on…..

Let’s investigate the four options.

A) 

\(T_{20} = 1604\)

\(\implies T_{19} + a_{19} = 1604\)

\(\implies T_{18} + a_{18} + a_{19} = 1604\)

……

\(\implies T_{1} + a_{1} + a_{2} + a_{3} + ….+ a_{19} = 1604\)

 \(\implies 3 + a_{1} + (a_{1} + d) + (a_{1} + 2d) + ….+ (a_{1} + 18d) = 1604\)

\(\implies 3 + (19 \times a_{1}) + ((1 + 2 + 3 + …+ 18) \times d) = 1604\)

\(\implies 3 + (19 \times 7) + \frac{18 \times 19 \times 8}{2} = 1604\)

\(\implies 1504 = 1604\)

which is not true.

C) 

Similarly, 

\(T_{30} = 3454\)

\(\implies T_{1} + a_{1} + a_{2} + ….+ a_{29} = 3454\)

\(\implies 3 + (29 \times a_{1}) + ((1 + 2 + …+ 28) \times d) = 3454\)

\(\implies 3 + (29 \times 7) + \frac{28 \times 29 \times 8}{2} = 3454\)

\(\implies 3454 = 3454\)

which is true.

B) 

\(\sum_{k=1}^{20} T_{k} = 10510\)

\(\implies T_{1} + T_{2} + T_{3} + T_{4} +….+ T_{20} = 10510\)

\(\implies T_{1} + (T_{1} + a_{1}) + (T_{1} + a_{1} + a_{2}) + (T_{1} + a_{1} + a_{2} + a_{3}) + ….+ (T_{1} + a_{1} + a_{2} +…. + a_{19}) = 10510\)

\(\implies 20T_{1} + 19a_{1} + 18a_{2} + ….+ 1a_{19} = 10510\)

Using \(a_{n} = a_{1} + (n - 1)d\)

\(\implies 20(3) + (19 + 18 + 17 +…+ 1)a_{1} + 18d + 17(2d) + 16(3d) …+ 2(17d) + 1(18d) = 10510\)

Let’s first focus on \(18d + 17(2d) + 16(3d) +.…+ 2(17d) + 1(18d)\)

\(= (19 - 1)(d) + (19 - 2)(2d) + (19 - 3)(3d) +….+ (19 - 17)(17d) + (19 - 18)(18d)\)

\(= 19d + 19(2d) + 19(3d) +….+ 19(18d) - [d + 2^{2}d + 3^{2}d +….+ 18^{2}d]\)

\(= 19d(1 + 2 +….+ 18) - [d(1^{2} + 2^{2} + 3^{2} +….+ 18^{2})]\)

\(= \frac{19 \times 8 \times 18 \times 19}{2} - \frac{8 \times 18 \times 19 \times 37}{6}\)

\(\implies 60 + \frac{19 \times 20 \times 7}{2} + \frac{19 \times 8 \times 18 \times 19}{2} - \frac{8 \times 18 \times 19 \times 37}{6} = 10510\)

\(\implies\) 60 + 1330 + 25992 - 16872 = 10510

\(\implies 10510 = 10510\)

which is true

D)

\(\sum_{k=1}^{30} T_{k} = 35610\)

\(\implies T_{1} + T_{2} + T_{3} + T_{4} +….+ T_{30} = 35610\)

\(\implies 30T_{1} + 29a_{1} + 28a_{2} + 27a_{3} + ….+ 1a_{29} = 35610\)

\(\implies 30T_{1} + (29 + 28 +.…+1)a_{1} + 28(d) + 27(2d) + ….+ 1(28d) = 35610\)

\(\implies 30(3) + \frac{29 \times 30 \times 7}{2} + (29 -1)(d) + (29 - 2)(2d) + ….+ (29 - 28)(28d) = 35610\)

\(\implies 90 + 3045 + 28(d) + 29d(1 + 2….+ 28) - d(1^2 + 2^2 +….+ 28^2) = 35610\)

\(\implies 90 + 3045 + 28(8) + \frac{29 \times 8 \times 28 \times 29}{2} - (\frac{8 \times 28 \times 29 \times 57}{6}) = 35610\)

\(\implies\) 90 + 3045 + 224 + 94192 - 61712 = 35610

\(\implies 35839 = 35610\)

which is not true

\(\implies\) Options B and C are True.


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