(i) given a = 5, d = 3, an = 50, find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2, Sn = –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, 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, the 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)
S13 = 13/2 (7 + 35) = 273
(iii) Given that, a12 = 37, d = 3
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,
⇒ a12 = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
Now, the sum of the nth term,
Sn = n /2 ( a + an )
Sn = 12 /2 (4 + 37)
= 246
(iv) Given that, a3 = 15, S10 = 125
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,
a 3 = a + (3 − 1) d
15 = a + 2d ………………………….. (i)
Sum of the n terms,
Sn = n /2 [2 a + (n – 1) d ]
S10 = 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
a10 = a + (10 − 1) d
a10 = 17 + (9)(−1)
a10 = 17 − 9 = 8
(v) Given that, d = 5, S9 = 75
As the sum of n terms in AP is,
Sn = n /2 [2 a + (n – 1) d ]
Therefore, the sum of the first nine terms is;
S9 = 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 nth term can be written as;
an = 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, Sn = 90
As the sum of n terms in an AP is,
Sn = n /2 [2 a + (n – 1) d ]
90 = n /2 [2 a + (n – 1) d ]
⇒ 180 = n (4 + 8n – 8) = n (8n – 4) = 8 n2 – 4n
⇒ 8n2 – 4n – 180 = 0
⇒ 2n2 – n – 45 = 0
⇒ 2 n2 – 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)
∴ a5 = 8 + 5 × 4 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
As the sum of n terms in an AP is,
Sn = n /2 ( a + an )
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 an = 4, d = 2, Sn = –14, find n and a.
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,
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;
Sn = n /2 ( a + an )
-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,
Sn = 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,
Sn = n /2 ( a + l )
144 = 9/2( a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4