# NCERT Class 11-Math՚s: Chapter – 9 Sequence and Series Part 6 (For CBSE, ICSE, IAS, NET, NRA 2023)

Doorsteptutor material for CBSE/Class-9 is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-9.

**Question 19**:

The 10^{th} common term between the series

and is

(A)

(B)

(C)

(D) None of these

**Answer**:

(A) Is the correct answer.

The first common term is 11.

Now the next common term is obtained by adding L. C. M. of the common difference 4 and , i.e.. , .

Therefore, common of the AP whose and

**Question 20**:

In a G. P. of even number of terms, the sum of all terms is times the sum of the odd terms. The common ratio of the G. P. is

(A)

(B)

(C)

(D) None the these

**Answer**:

(C) Is the correct answer

Let us consider a G. P. with terms. We have

(Since common ratio of odd terms will be and number of terms will be )

**Question 21**:

The minimum value of the expression , is

(A)

(B)

(C)

(D)

**Answer**:

(D) Is the correct answer.

We know . for positive numbers.

Therefore,

## 9.3 Exercise

**Short Answer Type**

**Question 1**:

The first term of an A. P. is *a*, and the sum of the first terms is zero, show that the sum of its next terms is .

**Answer**:

Given first term is ‘a’ and sum of first p terms is we have to find the sum of next terms

Total terms are

So, sum of all terms minus the sum of the first terms will give the sum of next terms

But sum of first terms is hence sum of next terms will be the same as sum of all terms

So, we have to find sum of terms

Sum of n terms of AP is given by

Where a is first term and d is the common difference

Using the given hint

Required

Using formula

Now we have to find d

We use the given to find

Put this value of d in (i)

Required sum

Hence proved