Question -
Answer -
(i) 50, 46, 42, …. to 10terms
n = 10
First term, a = a1 =50
Common difference, d =a2 – a1 = 46 – 50 = -4
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = 10/2 (100 + (9)(-4))
= 5 (100 – 36)
= 5 (64)
= 320
∴ The sum of the givenAP is 320.
(ii) 1, 3, 5, 7, … to 12terms
n = 12
First term, a = a1 =1
Common difference, d =a2 – a1 = 3 – 1 = 2
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = 12/2 (2(1) +(12-1) (2))
= 6 (2 + (11) (2))
= 6 (2 + 22)
= 6 (24)
= 144
∴ The sum of the givenAP is 144.
(iii) 3, 9/2, 6, 15/2, … to25 terms
n = 25
First term, a = a1 =3
Common difference, d =a2 – a1 = 9/2 – 3 = (9 – 6)/2 = 3/2
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = 25/2 (2(3) +(25-1) (3/2))
= 25/2 (6 + (24)(3/2))
= 25/2 (6 + 36)
= 25/2 (42)
= 25 (21)
= 525
∴ The sum of the givenAP is 525.
(iv) 41, 36, 31, … to 12terms
n = 12
First term, a = a1 =41
Common difference, d =a2 – a1 = 36 – 41 = -5
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = 12/2 (2(41) +(12-1) (-5))
= 6 (82 + (11) (-5))
= 6 (82 – 55)
= 6 (27)
= 162
∴ The sum of the givenAP is 162.
(v) a+b, a-b, a-3b, … to22 terms
n = 22
First term, a = a1 =a+b
Common difference, d =a2 – a1 = (a-b) – (a+b) = a-b-a-b = -2b
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = 22/2 (2(a+b) +(22-1) (-2b))
= 11 (2a + 2b + (21)(-2b))
= 11 (2a + 2b – 42b)
= 11 (2a – 40b)
= 22a – 440b
∴ The sum of the givenAP is 22a – 440b.
(vi) (x – y)2,(x2 + y2), (x + y)2, … to n terms
n = n
First term, a = a1 =(x-y)2
Common difference, d =a2 – a1 = (x2 + y2)– (x-y)2 = 2xy
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = n/2 (2(x-y)2 +(n-1) (2xy))
= n/2 (2 (x2 +y2 – 2xy) + 2xyn – 2xy)
= n/2 × 2 ((x2 +y2 – 2xy) + xyn – xy)
= n (x2 +y2 – 3xy + xyn)
∴ The sum of the givenAP is n (x2 + y2 – 3xy + xyn).
(vii) (x – y)/(x + y), (3x –2y)/(x + y), (5x – 3y)/(x + y), … to n terms
n = n
First term, a = a1 =(x-y)/(x+y)
Common difference, d =a2 – a1 = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x– y)/(x+y)
By using the formula,
S = n/2 (2a + (n – 1)d)
Substitute the valuesof ‘a’ and ‘d’, we get
S = n/2(2((x-y)/(x+y)) + (n-1) ((2x – y)/(x+y)))
= n/2(x+y) {n (2x-y) –y}
∴ The sum of the givenAP is n/2(x+y) {n (2x-y) – y}