Question -
Answer -
(i) 0.15 + 0.015 + 0.0015+ … to 8 terms
Given:
a = 0.15
r = t2/t1 =0.015/0.15 = 0.1 = 1/10
n = 8
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = 0.15 (1 – (1/10)8) / (1 – (1/10))
= 0.15 (1 – 1/108)/ (1/10)
= 1/6 (1 – 1/108)
(ii) √2 + 1/√2 + 1/2√2 + ….to 8 terms;
Given:
a = √2
r = t2/t1 =(1/√2)/√2 = 1/2
n = 8
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = √2 (1 – (1/2)8) / (1 – (1/2))
= √2 (1 – 1/256) /(1/2)
= √2 ((256 – 1)/256) ×2
= √2 (255×2)/256
= (255√2)/128
(iii) 2/9 – 1/3 + ½ – ¾ + …to 5 terms;
Given:
a = 2/9
r = t2/t1 =(-1/3) / (2/9) = -3/2
n = 5
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
a(1 – rn )/(1– r) = (2/9) (1 – (-3/2)5) / (1 – (-3/2))
= (2/9) (1 + (3/2)5)/ (1 + 3/2)
= (2/9) (1 + (3/2)5)/ (5/2)
= (2/9) (1 + 243/32) /(5/2)
= (2/9) ((32+243)/32)/ (5/2)
= (2/9) (275/32) × 2/5
= 55/72
(iv) (x + y) + (x2 +xy + y2) + (x3 + x2 y + xy2 +y3) + …. to n terms;
Let Sn =(x + y) + (x2 + xy + y2) + (x3 + x2 y+ xy2 + y3) + …. to n terms
Let us multiply anddivide by (x – y) we get,
Sn =1/(x – y) [(x + y) (x – y) + (x2 + xy + y2) (x – y)… upto n terms]
(x – y) Sn =(x2 – y2) + x3 + x2y + xy2 –x2y – xy2 – y3..upto n terms
(x – y) Sn = (x2 +x3 + x4+…n terms) – (y2 + y3 +y4 +…n terms)
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
We have two G.Ps inabove sum, so,
(x – y) Sn =x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]
Sn =1/(x-y) {x2 [(xn – 1)/ (x – 1)] – y2 [(yn –1)/ (y – 1)]}
(v) 3/5 + 4/52 +3/53 + 4/54 + … to 2n terms;
The series can bewritten as:
3 (1/5 + 1/53 +1/55+ … to n terms) + 4 (1/52 + 1/54 +1/56 + … to n terms)
Firstly let usconsider 3 (1/5 + 1/53 + 1/55+ … to n terms)
So, a = 1/5
r = t2/t1 =1/52 = 1/25
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
Now, Let us consider 4(1/52 + 1/54 + 1/56 + … to nterms)
So, a = 1/25
r = t2/t1 =1/52 = 1/25
By using the formula,
Sum of GP for n terms= a(1 – rn )/(1 – r)
