Question 1 Find the sum of the following APs. (i) 2, 7, 12 ,…., to 10 terms. (ii) − 37, − 33, − 29 ,…, to 12 terms (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms (iv) 1/15, 1/12, 1/10,………, to 11 terms
Class 10 - Math - Arithmetic Progressions Page 112
Answers
(1) a = 2 , d = 7 - 2 = 5 and n = 10
S10 = 10/2{ 2×2 + (10 -1)× 5}
= 5 { 4 + 45}
= 5 × 49
= 245
(ii) a = -37 , d = 4 and n = 12
S12 = 12/2{ -37×2 + (12-1)×4}
= 6{ -74 + 44}
= -180
(iii) a = 0.6 , d = 1.1 and n = 100
S100 = 100/2{ 2 × 0.6 + (100-1)×1.1 }
= 50 × { 1.2 + 108.9}
= 50 × 110.1
= 5505
(iv) a = 1/15 , d = 1/12 - 1/15 = 1/60
S11 = 11/2{ 2/15 + 10 × 1/60 }
= 11/2 { 18/60 }
= 33/20
Sum of n terms of an AP
The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by
Sn = n /2 [ 2a + ( n - 1) d] or
Sn=n /2 [ a + l ] (l = an= last term)
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(i) 2, 7, 12 ,…, to 10 terms
Given:
a = 2
d = a2 − a1 = 7 − 2 = 5
n = 10
We know that,
Sn = n/2 [2a + (n – 1) d]
S10 = 10/2 [2(2) + (10 – 1) × 5]
⇒ 5[4 + (9) × (5)]
⇒ 5[4+45]
⇒ 5 × 49 = 245
S10= 245
(ii) −37, −33, −29 ,…, to 12 terms
Given:
a = −37
d = a2 − a1
⇒ (−33) − (−37)
⇒ − 33 + 37 = 4
d= 4
n = 12
We know that,
Sn = n/2 [2a + (n – 1) d]
S12 = 12/2 [2(-37) + (12 – 1) × 4]
⇒ 6[-74 + 11 × 4]
⇒ 6[-74 + 44]
⇒ 6(-30) = -180
S12 = -180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
Given:
a = 0.6
d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that,
Sn = n/2 [2a + (n – 1) d]
S100 = 100/2 [1.2 + (99) × 1.1]
⇒ 50[1.2 + 108.9]
⇒ 50[110.1] = 5505
S100= 5505
(iv) 1/15, 1/12, 1/10, …… , to 11 terms
Given:
a= 1/15
d= 1/12- 1/15
d=(5-4) / 60 = 1/60
d= 1/60
n= 11
Sn = n/2 [2a + (n – 1) d]
S11= 11/2 [ 2×1/15 + (11-) 1/60]
⇒ 11/2( 2/15 + 10 ×1/60)
⇒11/2(2/15 +1/6)
⇒11/2 ×(4+5/30)
⇒11/2 ×9/30= 33/20
S11= 33/20
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Hope this will help you....