Question

Question 2 Find the sums given below (i) 7 + 10.1/2 + 14 + ………… + 84 (ii) 34 + 32 + 30 + ……….. + 10 (iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Class 10 - Math - Arithmetic Progressions Page 112

Answer 1

AP  ( Arithmetic progression).
 
A list of numbers a1, a2, a3  ………….. an is called an arithmetic progression ,

Is there exists a constant number ‘d’
a2= a1+d
a3= a2+d
a4= a3+d……..
an= an-1+d ………

Each of the numbers in the list is called a term .

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

This fixed  number is called the common difference( d ) of the AP.

General form of an AP.:
a, a+d, a+2d, a+3d…….

Here a is the first term and d is common difference.

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 = last term)

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Solution:

(i)Given:

a = 7
l = an= 84
d = a2 − a1 =   – 7 = 21/2 – 7 = 7/2
d= 7/2

Let 84 be the nth term of this A.P.
l = a (n – 1)d
84 = 7 + (n – 1) × 7/2
77 = (n – 1) × 7/2
22 = n − 1
n = 23

We know that,
Sn = n/2 (a + l)

Sn = 23/2 (7 + 84)
      = (23×91/2) = 2093/2

 Sn=  1046 1/2

(ii) 34 + 32 + 30 + ……….. + 10
Given:
a = 34
d = a2 − a1 = 32 − 34 = −2d= -2l = an= 10

Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13

Sn = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22

Sn= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230)

Given:
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
d= -3

Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76

Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)] = 38×(-235)

Sn= -8930
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Hope this will help you....

Answer 2

(i) For this A.P.,
a = 7
l = 84
d = a2 − a1 =10×1/2 - 7 = 21/2 - 7 = 7/2
Let 84 be the nth term of this A.P.
l = a (n - 1)d
84 = 7 + (n - 1) × 7/2
77 = (n - 1) × 7/2
22 = n − 1 
n = 23
We know that,
Sn = n/2 (a + l)
Sn = 23/2 (7 + 84)
= (23×91/2) = 2093/2
= 1046.1/2


(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a2 − a1 = 32 − 34 = −2
l = 10
Let 10 be the nth term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n = 13
Sn = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,
a = −5
l = −230
d = a2 − a1 = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the nth term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And, 
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930