Find the sums given below :(i) 7 + 10 1/2 + 14 .... + 84(iii) 34 + 32 + 30 + . . . + 10
(iii) –5 + (–8) + (–11) + . . . + (–230)
Answers
Now We are asked to find sum of some number of terms in Arithmetic progression.
We know that, Sum of n terms = n/2 ( a + l)
where a is the first term, l is the last term.
(i) 7 + 10 1/2 + 14 + ... + 84
Here a = 7
l = 84
d = 3.5
We know that, nth term of this A. P is l
84 = a + ( n - 1 ) d
84 = 7 + ( n - 1 ) 3.5
84 = 7 - 3.5 + 3.5n
84 - 3.5 = 3.5n
80.5/3.5 = n
n = 23 .
So sum of first 23 terms = 7 + 10 1/2 + 14 ... + 84 = 23/2 ( 7 + 84 )
= 11.5 * 91
= 1046.5
(ii) 34 + 32 + 30 + . . . + 10
First term ( a) = 34
Last term ( l) = 10
Common difference ( d) = 32 - 34 = -2
We know that, the nth term of the A. P is l
10 = 34 + ( n - 1 )d
10 = 34 + (n-1)*-2
10 - 34 = -2n + 2
-24 - 2 = -2n
-26 = -2n .
n = 13
Sum of first thirteen terms = 34 + 32 + 30+ ... + 10 = 13/2 ( 34 + 10 ) = 13 * 22 = 286 .
(iii) –5 + (–8) + (–11) + . . . + (–230)
First term( a) = -5
Last term ( l) = -230
Common difference = -8 - ( -5 ) = -3
Let the nth term of this A. P be l
-230 = -5 + ( n - 1 ) d
-230 + 5 = ( n - 1 )* -3
-225 = -3n + 3
-225- 3 = -3n
-228 = -3n
n = -228 / -3 = 76
Sum of 76 terms = –5 + (–8) + (–11) + . . . + (–230) = 76/2 ( -5 -230 ) = 38* -235 = -8930
Conclusion :
(i) 7 + 10 1/2 + 14 .... + 84 = 1046.5
(ii) 34 + 32 + 30 + . . . + 10 = 286
(iii) –5 + (–8) + (–11) + . . . + (–230) = -8930
Hope helped!
Answer:
Step-by-step explanation:
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