2 + 4 + 6 + ......... to n terms
This is an arithmetic sequence with t1 = 2 and d = 2.
Formula for sum of first n terms of an arithmetic sequence :
Sn = (n/2)[2t1 + (n - 1)d]
Substitute a1 = 2 and d = 2.
= (n/2)[2(2) + (n - 1)2]
= (n/2)[4 + 2n - 2]
= (n/2)[2n + 2]
= (n/2) ⋅ 2(n + 1)
= n(n + 1)
2 + 4 + 6 + ........ to n terms = n(n + 1)
In the sum of first n even natural numbers, if the number of terms is not given and the last term l is given, then formula for finding number of terms n :
n = l/2
Example 1 :
Find the sum of
2 + 4 + 6 + ........ to 50 terms
Solution :
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
2 + 4 + 6 + ........ + 50 terms = 50(50 + 1)
= 50(51)
= 2550
Example 2 :
Find the value of
2 + 4 + 6 + ........ + 90
Solution :
Find the value of n.
n = l/2
= 90/2
= 45
2 + 4 + 6 + ........ + 90 :
= 2 + 4 + 6 + ........ to 45 terms
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
= 45(45 + 1)
= 45(46)
= 2070
Example 3 :
Find the value of
22 + 24 + 26 + ........ + 100
Solution :
22 + 24 + 26 + ........ + 100 :
= (2 + 4 + 6 + ........ + 100) - (2 + 4 + 6 + ........ + 20)
= (2 + 4 + 6 + .... to 50 terms) - (2 + 4 + 6 + .... to 10 terms)
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
= 50(50 + 1) - 10(10 + 1)
= 50(51) - 10(11)
= 2550 - 110
= 2440
Example 4 :
Using the sum of first n even natural numbers, find the sum of
6 + 12 + 18 + ........ to 50 terms
Solution :
6 + 12 + 18 + ........ to 50 terms :
= 3(2 + 4 + 6 + ........ to 50 terms)
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
= 3[50(50 + 1)]
= 3[50(51)]
= 7650
Example 5 :
Using the sum of first n even natural numbers, find the value of
6 + 12 + 18 + ........ + 360
Solution :
6 + 12 + 18 + ........ + 360 :
= 3(2 + 4 + 6 + ........ + 120)
= 3(2 + 4 + 6 + ........ to 60 terms)
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
= 3[60(60 + 1)]
= 3[60(61)]
= 10980
Example 6 :
If 2 + 4 + 6 + ........ + k = 72, then find k.
Solution :
Find the value of n.
n = l/2
= k/2
2 + 4 + 6 + ........ to k/2 terms = 72
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
(k/2)(k/2 + 1) = 72
(k/2)[(k + 2)/2] = 72
k(k + 2)/4 = 72
Multiply each side by 4.
k(k + 2) = 288
k2 + 2k = 288
k2 + 2k - 288 = 0
Factor and solve.
k2 + 18k -16k - 288 = 0
k(k + 18) - 16(k + 18) = 0
(k + 18)(k - 16) = 0
k + 18 = 0 k = -18 |
k - 16 = 0 k = 16 |
But k ≠ -18, because k is an even natural number.
Hence k = 16.
Example 7 :
If 2 + 4 + 6 + ........ to k terms = 240, then find k.
Solution :
2 + 4 + 6 + ........ to k terms = 240
Using 2 + 4 + 6 + ........ to n terms = n(n + 1),
k(k + 1) = 240
k2 + k = 240
k2 + k - 240 = 0
Factor and solve.
k2 + 16k - 15k - 240 = 0
k(k + 16) - 15(k + 16) = 0
(k + 16)(k - 15) = 0
k + 16 = 0 k = -16 |
k - 15 = 0 k = 15 |
But k ≠ -16, because number of terms can not be negative value.
Hence k = 15.
Example 8 :
Find the average of first 25 even natural numbers.
Solution :
Using 2 + 4 + 6 + ........ to n terms = n(n + 1) to find the sum of first 25 odd natural numbers.
2 + 4 + 6 + ........ to 25 terms = 25(25 + 1)
= 25(26)
= 650
Average of first 25 even natural numbers :
= (Sum of first 25 even natural numbers)/25
= 650/25
= 26
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