Solve the equation for x: 72 + 70 + 68 +.......+x=952
this is arithematic progression sum
Answers
Answer:
Formulae for Sum of A.P.
Formulae
Sum of first n terms of an A.P. is
Sn = (n/2)[2a +(n -1)d]
If Sn Is sum of n terms of an A.P. whose first term is a and last term is l,
then Sn = (n/2)(a + l)
If common difference is d, number of terms n and the last term l,
then Sn = (n/2)[2l-(n -1)d]
Remark
Sometimes the question involves 3, 4 or 5 terms of an A.P.
If the sum of the numbers is given, then in an A.P.,
three numbers are taken as a -d, a, a +d
four numbers are taken as a -3d, a -d, a +d, a +3d.
five numbers are taken as a -2d, a -d, a, a +d, a +2d.
This considerably simplifies the calculations of a and d.
Illustrative Examples
Example
Sum the series 2 +4 +6 +... upto 40 terms.
Find the sum of first 19 terms of the A.P. whose nth term is 2n +1.
Find the sum of the series 1 +3 +5 +... +99.
Solution
We see that given series is an A.P. with first term a = 2,
common difference d = 2 and number of terms n = 40
Hence S40 = n[2a +(n -1)d]/2 = 40[2.2 +(40 -1)2]/2
= 20(4 +78) = 1640
Here first term a = T1 = 2n +1 = 2(1) +1 = 3,
and last term = T19 = 2(19) +1 = 39
S19 = n(a +l)/2 = 19(3 +39)/2 = 19.21 = 399
Given series is an A.P. with a = 1, l = 99, d = 2
To find the number of terms, we use l = a +(n -1)d
=> 99 = 1 +(n -1)2 => n -1 = 49 => n = 50
Sn = n(a +l)/2 = 50(1 +99)/2 = 50.50 = 2500
Example
Sum up 0·7 +0·71 +0·72 +... upto 100 terms.
How many terms of the A.P. 17 +15 +13 +... must be taken so that sum is 72? Explain the double answer.
Solution
We see that given series is an A.P. with first term
a = 0·7 and common difference d = 0·01
S100 = n[2a +(n -1)d]/2 = 100[2(0·7) +(100 -1)(0·01)]/2
= 50 (1· 4 +0·99) = 119·5
We are given a = 17, d = -2, Sn = 72 and we have to find n.
Using Sn = n[2a +(n -1)d]/2, we get
72 = n[2.17 +(n -1)(-2)]/2 = n(36 -2n)/2 = n(18 -n)
=> n² -18n +72 = 0 => (n -6)(n -12) = 0
=> n = 6, 12
Both values of n, being positive integers are valid. We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative.
Example
The interior angles of a polygon are in arithmetic progression. The smallest angle is 120° and the common difference is 5°. Find the number of sides in the polygon.
Solution
The sum of all exterior angles of a polygon = 360°. If the interior angles are in A.P., then exterior angles are also in A.P.; largest exterior angle is 180°-120° = 60° and common difference is -5°.
Sum = 360° = n [2 a +(n -1) d]/2 = [2. 60° +(n -1)(-5°)]/2
=> 720 = 120 n -5n² +5n => 5n² -125n +720 = 0
=> n² -25 n +144 = 0 => (n -9)(n -16) = 0
=> n = 9 or n = 16.
However if n = 16, then internal angles would vary from 120° to 120° +(16 -1)(5°) = 195°; so one of the internal angles will be 180° which is not possible in a polygon.
Hence the only correct solution is n = 9.
Example
The sum of three numbers in A.P. is 18 and their product is 192. Find the numbers.
Solution
Let the three numbers in A.P. be a -d, a, a +d. By given conditions,
(a -d) +a +(a +d) = 18, (a -d)a(a +d) = 192
=> 3a = 18, a(a² -d²) = 192 => a = 6, 6(36 -d²) = 192
=> d² = 36 -192/6 = 36 -32 = 4 => d = ±2.
With a = 6, d = 2, we get three numbers as 4, 6, 8.
With a = 6, d = -2, we get three numbers as 8, 6, 4.
Hence the required numbers are 4, 6, 8.