(i) given a = 5, d = 3, a_{n} = 50, find n and S_{n}.

(ii) given a = 7, a_{13} = 35, find d and S_{13}.

(iii) given a_{12} = 37, d = 3, find a and S_{12}.

(iv) given a_{3} = 15, S_{10} = 125, find d and a_{10}.

(v) given d = 5, S_{9} = 75, find a and a_{9}.

(vi) given a = 2, d = 8, S_{n} = 90, find n and a_{n}.

(vii) given a = 8, a_{n} = 62, S_{n} = 210, find n and d.

(viii) given a_{n} = 4, d = 2, S_{n} = –14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

**Solution:**

**(i) Given that, a = 5, d = 3, a _{n} = 50**

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

a_{n} = 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, the sum of n terms,

S_{n} = n /2 (a + a_{n})

S_{n} = 16/2 (5 + 50) = 440

**(ii) Given that, a = 7, a _{13} = 35**

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

a_{n} = 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, S_{n} = n /2 (a + a_{n})

S_{13} = 13/2 (7 + 35) = 273

**(iii) Given that, a _{12} = 37, d = 3**

As we know, from the formula of the n^{th} term in an AP,

a_{n} = a + (n − 1) d ,

Therefore, putting the given values, we get,

⇒ a_{12} = a + (12 − 1)3

⇒ 37 = a + 33

⇒ a = 4

Now, the sum of the nth term,

S_{n} = n /2 ( a + a_{n} )

S_{n} = 12 /2 (4 + 37)

= 246

**(iv) Given that, a _{3} = 15, S_{10} = 125**

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

a_{n} = a + (n − 1) d ,

Therefore, putting the given values, we get,

a _{3} = a + (3 − 1) d

15 = a + 2d ………………………….. **(i)**

Sum of the n terms,

S_{n} = n /2 [2 a + (n – 1) d ]

S_{10} = 10/2 [2 a + (10 – 1) d ]

125 = 5(2 a + 9 d )

25 = 2 a + 9 d ……………………….. **(ii)**

On multiplying equation **(i)** by **(ii)**, we will get;

30 = 2 a + 4 d ………………………………. **(iii)**

By subtracting equation **(iii)** from **(ii)**, we get,

−5 = 5 d

d = −1

From equation **(i)**,

15 = a + 2(−1)

15 = a − 2

a = 17 = First term

a_{10} = a + (10 − 1) d

a_{10} = 17 + (9)(−1)

a_{10} = 17 − 9 = 8

**(v) Given that, d = 5, S _{9} = 75**

As the sum of n terms in AP is,

S_{n} = n /2 [2 a + (n – 1) d ]

Therefore, the sum of the first nine terms is;

S_{9} = 9/2 [2 a + (9 – 1) 5 ]

25 = 3(a + 20)

25 = 3 a + 60

3 a = 25 − 60

a = -35/3

As we know, the n^{th} term can be written as;

a_{n} = a + (n − 1) d

a _{9} = a + (9 − 1)(5)

= -35/3 + 8(5)

= -35/3 + 40

= (35 + 120/3) = 85/3

**(vi) Given that, a = 2, d = 8, S _{n} = 90**

As the sum of n terms in an AP is,

S_{n} = n /2 [2 a + (n – 1) d ]

90 = n /2 [2 a + (n – 1) d ]

⇒ 180 = n (4 + 8n – 8) = n (8n – 4) = 8 n^{2 }– 4n

⇒ 8n^{2 }– 4n – 180 = 0

⇒ 2n^{2 }– n – 45 = 0

⇒ 2 n^{2 }– 10n + 9n – 45 = 0

⇒ 2n (n – 5) + 9( n – 5) = 0

⇒ (n – 5)(2n + 9) = 0

So, n = 5 (as n only be a positive integer)

∴ a_{5} = 8 + 5 × 4 = 34

**(vii) Given that, a = 8, a _{n} = 62, S_{n} = 210**

As the sum of n terms in an AP is,

S_{n} = n /2 ( a + a_{n} )

210 = n /2 (8 + 62)

⇒ 35 n = 210

⇒ n = 210/35 = 6

Now, 62 = 8 + 5 d

⇒ 5 d = 62 – 8 = 54

⇒ d = 54/5 = 10.8

**(viii) given a _{n} = 4, d = 2, S_{n} = –14, find n and a.**

As we know, from the formula of the n^{th} term in an AP,

a_{n} = a + (n − 1) d ,

Therefore, putting the given values, we get,

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n …………………………………………. **(i)**

As we know, the sum of n terms is;

S_{n} = n /2 ( a + a_{n} )

-14 = n /2 ( a + 4 )

−28 = n ( a + 4)

−28 = n (6 −2 n + 4) {From equation **(i)**}

−28 = n (− 2 n + 10)

−28 = − 2 n ^{2 }+ 10 n

2 n ^{2} − 10 n − 28 = 0

n ^{2} − 5 n − 14 = 0

n ^{2} −7 n + 2 n − 14 = 0

n ( n − 7) + 2( n − 7) = 0

( n − 7)( n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation **(i)**, we get

a = 6 − 2 n

a = 6 − 2(7)

= 6 − 14

= −8

**(ix) given a = 3, n = 8, S = 192, find d.**

Number of terms, n = 8

And the sum of n terms, S = 192

As we know,

S_{n} = n /2 [2 a + ( n – 1) d ]

192 = 8/2 [2 × 3 + (8 – 1) d ]

192 = 4[6 + 7 d ]

48 = 6 + 7 d

42 = 7 d

d = 6

**(x) Given that, l = 28, S = 144 and there are total of 9 terms.**

The sum of n terms formula,

S_{n} = n /2 ( a + l )

144 = 9/2( a + 28)

(16) × (2) = a + 28

32 = a + 28

a = 4