In an AP:

(i) givenfind and

(ii) givenfind and

(iii) given find and

(iv) givenfind and

(v) givenfind a and

(vi) givenfind and

(vii) given find and

(viii) given find and

(ix) given find

(x) givenand there are total 9 terms. Find

(i) Number of terms can be calculated as follows:

a_n = a + (n – 1)d

Or, 50 = 5 + (n – 1)3

Or, (n – 1)3 = 50 – 5 = 45

Or, n – 1 = 15

Or, n = 16

Sum of n terms can be given as follows:

S₁₆

= (10 + 45)

=

=440

(ii) Common difference can be calculated as follows:

a_n = a + (n – 1)d

Or, 35 = 7 + 12d

Or, 12d = 35 – 7 = 28

Or, d = 7/3

Sum of n terms can be given as follows:

S₁₃

= (14 + 28)

=

=273

(iii) First term can be calculated as follows:

a_n = a + (n – 1)d

Or, 37 = a + 11 x 3

Or, a = 37 – 33 = 4

Sum of n terms can be given as follows:

S₁₂

= 6(8 + 33)

= 6 * 41

= 246

(iv) Sum of n terms can be given as follows:

S₁₀

125 = 5(2a + 9d)

25 = 2a * 9d (i)

According to the question; the 3rd term is 15, which means;

a + 2d = 15 (ii)

Now,

Subtracting equation (ii) from equation (i), we get;

2a + 9d – a – 2d = 25 – 15

Or, a + 7d = 10 (iii)

Subtracting (ii) from equation (iii), we get;

a + 7d – a – 2d = 10 – 15

Or, 5d = - 5

Or, d = - 1

Now,

Substituting the value of d in equation (2), we get;

a + 2(- 1) = 15

Or, a – 2 = 15

Or, a = 17

10^th term can be calculated as follows;

a₁₀ = a + 9d

= 17 – 9 = 8

Thus, d = - 1 and 10th term = 8

(v) Sum of n terms can be given as follows:

75

75= (2a + 40)

= 2a * 40

Now,

9^th term can be calculated as follows:

a₉ = a + 8d

= -

=

(vi) Sum of n terms can be given as follows:

90

90= (4 + 8N -8)

90 = N(2 +4N -4)

4N² -2N -90 =0

2N² -N -45 =0

(2N +9)(N -5)

Hence, n = - 9/2 and n = 5

Rejecting the negative value,

We have n = 5

Now, 5^thterm will be:

a₅ = a + 4d

= 2 + 4 x 8

= 2 + 32 = 34

(vii) Sum of n terms can be given as follows:

210

420 = (8 + 62)

420 = n *70

n = 6

Now, for calculating d:

a₆ = a + 5d

Or, 62 = 8 + 5d

Or, 5d = 62 – 8 = 54

Or, d = 54/5

(viii) Sum of n terms can be given as follows:

-14

-28 = (a + 4)

(i)

We know;

a_n = a + (n – 1)d

Or, 4 = a + (n – 1)2

Or, 4 = a + 2n – 2

Or, a + 2n = 6

Or, 2n = 6 – a

Or, n = (6 – a)/2 (ii)

Using (i) and (ii):

-56 = (6 –a)(a +4)

24 + 2a – a² = -56

a² -2a -80= 0

(a + 8)(a – 10)= 0

Therefore, a = - 8 and a = 10

As a is smaller than 10 and d has positive value, hence we’ll take a = - 8

Now, we can find the number of terms as follows:

a_n = a + (n – 1)d

4 = - 8 + (n – 1)2

(n – 1)2 = 4 + 8 = 12

n – 1 = 6Or, n = 7

Hence, n = 7 and a = - 8

(ix) Sum of n terms can be given as follows:

192

192 = (6 + 7d)

7d = 42

d = 6

(x) Sum of n terms can be given as follows:

144 (a + a_n)

288 = (a + 28)

9a + 252 = 288

9a = 288 -252

9a = 36

a = 4