Question

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Answer 1

Consider,

S = 3 + 5 + 8 + 12 + 17 + …………………………………….. + T(n)

Shift by one position,

S = 0 + 3 + 5 + 8 + 12 + 17 +……………………………….+ T(n-1) + T(n)

- - - - - -←——————————N TERMS—————————————→

Subtract both equations,

0 = 3 + 2 + 3 + 4 + 5 +……………………………………………+ (T(n)-T(n-1))-T(n)

- - - - - - ←———————————N TERMS————————————->

T(n) = 3 + 2 + 3 + 4 + 5 +……………………………………………+ (T(n)-T(n-1))

- - - - - - - - ←————————————-(N-1) TERMS——————>

T(n) - 3 = 2 + 3 + 4 + 5 + ………………………………………………………..+ F(n-1)

F(x) is different from T(x)

T(n) is the nth term of the given sequence

F(n) is the nth term of the new Arithmetic sequence having 2 as first number and 1 as the common difference

Now sum of an arithmetic progression is (n(2×a+(n-1)d))÷2

i.e (no. of terms(2×first term + (no. of terms -1)×common difference))÷2

Sum till F(n-1) will be,

S(n-1) = (n-1(4+(n-2)×1))÷2

S(n-1)= (n-1)(n+2)÷2

T(n) = 3 + (n-1)(n+2)÷2

T(1) = 3 + 0 = 3

T(2) = 3 + 2 = 5

T(3) = 3 + 5 = 8

T(4) = 3 + 9 = 12

T(5) = 3 + 14 = 17

T(6) = 3 + 20 = 23

Hence,

Missing term is 23

Answer 2

Answer:

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