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# RD Chapter 21- Some Special Series Ex-21.2 Interview Questions Answers

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Question 1 : Sum the following series to n terms:

3 + 5 + 9 + 15 + 23 + ………….

Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.

We have,

Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn … (1)

Equation (1) can berewritten as:

Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn ……..(2)

By subtracting (2)from (1) we get

Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn

Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn

0 = 3 + [2 + 4 + 6 + 8+ … + (Tn – Tn-1)] – Tn

The difference betweenthe successive terms are 5-3 = 2, 9-5 = 4, 15-9 = 6,

So these differencesare in A.P

Now,

The sum of the seriesis n/3 (n2 + 8)

Question 2 :

2 + 5 + 10 + 17 + 26 + ………..

Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.

We have,

Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn … (1)

Equation (1) can berewritten as:

Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn ……..(2)

By subtracting (2)from (1) we get

Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn

Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn

0 = 2 + [3 + 5 + 7 + 9+ … + (Tn – Tn-1)] – Tn

The difference betweenthe successive terms are 3, 5, 7, 9

So these differencesare in A.P

Now,

The sum of the seriesis n/6 (2n2 + 3n + 7)

Question 3 :

1 + 3 + 7 + 13 + 21 + …

Let Tn bethe nth term and Sn be the sum to n terms of the given series.

We have,

Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn … (1)

Equation (1) can berewritten as:

Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn ……..(2)

By subtracting (2)from (1) we get

Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn

Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn

0 = 1 + [2 + 4 + 6 + 8+ … + (Tn – Tn-1)] – Tn

The difference betweenthe successive terms are 2, 4, 6, 8

So these differencesare in A.P

Now,

The sum of the seriesis n/3 (n2 + 2)

Question 4 :

3 + 7 + 14 + 24 + 37 + …

Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.

We have,

Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn … (1)

Equation (1) can berewritten as:

Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn ……..(2)

By subtracting (2)from (1) we get

Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn

Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn

0 = 3 + [4 + 7 + 10 +13 + … + (Tn – Tn-1)] – Tn

The difference betweenthe successive terms are 4, 7, 10, 13

So these differencesare in A.P

Now,

The sum of the seriesis n/2 [n2 + n + 4]

Question 5 :

1 + 3 + 6 + 10 + 15 + …

Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.

We have,

Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn … (1)

Equation (1) can berewritten as:

Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn ……..(2)

By subtracting (2)from (1) we get

Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn

Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn

0 = 1 + [2 + 3 + 4 + 5+ … + (Tn – Tn-1)] – Tn

The difference betweenthe successive terms are 2, 3, 4, 5

So these differencesare in A.P

Now,

The sum of the seriesis n/6 (n+1) (n+2)

krishan