**1- Write first four terms of the A.P. when the first term a and the common difference are given as follows:**

**(i) ***a* = 10, *d* = 10

**(ii) ***a* = -2, *d* = 0

**Solutions:**

**(i) ***a* = 10, *d* = 10

Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

**2- Which term of the A.P. 3, 8, 13, 18, … is 78?**

**Solutions:**

Given the A.P. series as3, 8, 13, 18, …

First term, a = 3

Common difference, d = a2 − a1 = 8 − 3 = 5

Let the *n*th term of given A.P. be 78. Now as we know,

*an = a+(n−1)d*

Therefore,

78 = 3+(*n* −1)5

75 = (*n*−1)5

(*n*−1) = 15

*n = 16*

Hence, the 16th term of this A.P. is 78.

**3- If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.**

**Solution:**

Given that,

3rd term, a3 = 4

and 9th term, a9 = −8

We know that,

an = a+(n−1)d

Therefore,

a3 = a+(3−1)d

4 = a+2d ……………………………………… **(i)**

a9 = a+(9−1)d

−8 = a+8d ………………………………………………… **(ii)**

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation **(i)**, we can write,

4 = *a*+2(−2)

4 = *a*−4

*a* = 8

Let the nth term of this A.P. be zero.

*an = a+(n−1)d*

0 = 8+(*n*−1)(−2)

0 = 8−2*n*+2

2*n* = 10

*n* = 5

Hence, the 5th term of this A.P. is 0.

4- **In an AP**

**(i) Given a = 5, d = 3, an = 50, find n and Sn.**

**(ii) Given a = 7, a13 = 35, find d and S13.**

** **

**Solutions:**

**(i) Given that, a = 5, d = 3, an = 50**

As we know, from the formula of the nth term in an AP,

*an = a +(n −1)d,*

Therefore, putting the given values, we get,

⇒ 50 = 5+(*n* -1)×3

⇒ 3(*n* -1) = 45

⇒ *n* -1 = 15

⇒ *n* = 16

Now, sum of n terms,

*Sn* = *n*/2 (*a +an*)

*Sn = 16/2 (5 + 50) = 440*

**(ii) Given that, a = 7, a13 = 35**

As we know, from the formula of the nth term in an AP,

*an = a+(n−1)d,*

Therefore, putting the given values, we get,

⇒ 35 = 7+(13-1)*d*

⇒ 12*d *= 28

⇒ *d* = 28/12 = 2.33

Now, *Sn = n*/2 (*a+an*)

S*13 = 13/2 (7+35) = 27*

**5- The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx]**

**Solution:**

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First-term, a = 1

The Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = Sx-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write,

S49 – Sx = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other;

Therefore,

x(x-1)/2 = 25(49) – x(x+1)/2

x = ±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.