if the nth term of an A. P is 4n-7,find the sum of the first 40 terms.
Answers
Answer:
“n”th term is given to be 3n + 7.
First term = 3*1 + 7 = 10;
Second term = 3*2 + 7 = 13;
Third term = 3*3 + 7 = 16;
Fourth term = 3*4 + 7 = 19;
The arithmetic progression is, therefore, 10, 13, 16, 19,…. with the first term a = 10, and common difference d = 3.
Sum of the first “n” terms of an arithmetic progression is given by
S = [n/2]*{2*a + (n-1)d}.
Sum of the first 40 terms of the arithmetic progression with “a” = 10 and “d” = 3 is
S = {40/2]{2*10 + (40–1)*3} = 20*{20 + 39*3} = 20*{20 + 117} = 20 * 137 = 2740
Sum of the first forty terms = 2740.
Alternately,
Sum of the first “n” terms of the series whose “n”th term is 3*n + 7 will be
3*(1+2+3+4+5+…..+n) + (7+7+7+7+….n times) = 3* n(n+1)/2 + 7n = n* [3(n+1)/2 + 7] = n(3n + 17)/2.
Hence, the sum of the first 40 terms will be 40 * (3*40 + 17)/2 = 40*137 / 2 = 2740.
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What is the sum of 10 terms of an arithmetic sequence whose first term is 5 and the last term is 40?
nth term =3n+7
1st term 3×1+7=10
2nd term=3×2+7= =13
3rd term=3×3+7=16
16–13=13–10=3
Therefore a= first term=10
d= common difference =3
Sum of nth term=(n/2) {2a+(n-1)d}
Substituting
=(40/2){2×10+(40–1)10}
20(20+390)
=20×410
=820
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3(1)+7
+ 3(2)+7
+ 3(3)+7
+ 3(4)+7
+ 3(5)+7
+ 3(6)+7
.
.
.
+ 3(38)+7
+ 3(39)+7
+ 3(40)+7
= 3(1+2+3…+38+39+40) + 7(40)
so what is the sum of the integers from 1 to 40? sum from 1 to n = (1/2) [n(n+1)]
sum from 1 to 40 is (1/2)[40(41)] = 20×41 = 820
so we have 3(820) + 7(40) = 2460 +280 = 2740
an=3n+7
a1=3(1)+7=10
a2=3(2)+7=13
a=10,d=13−10=3
sum=n2(2a+(n−1)d)=402(2(10)+(40−1)(3))
=20(20+(39)(3))=20(137)=2740
Sum of the first 40 terms is equal to 2740
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The mean of a group of numbers is the sum of all the numbers divided by how many numbers there are. Turning this around, the sum of a group of numbers is the mean multiplied by how many numbers there are. In an arithmetic progression, the mean can be calculated as the mean of the highest and lowest number in the series, which in turn can be calculated by adding them together and dividing by two.
In this question, the value of the first term is 10 (where n=1) and the value of the fortieth is 127. Therefore the mean is 68.5 and thus the sum is 68.5 X 40 = 2740.
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Let’s do this in our head as young Carl Friedrich Gauß did a similar exercise.
The first (assuming we start with n=1) term is 10, the 40th 127, their sum is 137. Obviously the sum of the second term and the 39th is the same and so on …
So we have 20 pairs, their sum is 20 * 137 = 2740.
PS:
Gauß and the others in his class were asked to add the integers from 1 to 100 and while the others were furiously working he thought for a minute and then wrote down the answer - his teacher thought he was too lazy to do the calculation …
The nth term of an AP is 3n+7
So T1 = 10, T2 = 13, T3 = 16 or
a = 10 and d = 3
So the sum of the first 40 terms is
S40 = (40/2)[2*10 + (40–1)*3]
= 20[20+117]
= 20*137
= 2740. Answer.
nth term is 3n+7 .
1st term =3×1+7=3+7=10
40th term=3×40+7=120+7=127
Let sum of the first 40 terms=S
so S=40/2(10+127)=20×137=2740
According to the given condition,
1st term =3*1+7=3+7=10
40th term=3*40+7=120+7=127
Total terms =40
Sum ={(1st term+last term)*Total terms}/2.
Sum= (10+127)*40/2
=137*40/2=137*20=2740.
So the answer is 2740.
Thanks.
n-1 th term = 3(n-1) +7
= 3n +4
d = n th term - (n-1) th term
= 3n +7 - 3n -4 = 3
first term = 3(1) + 7 = 10
So sum of 40 terms = 20 ( 20 + 39×3) = 20( 20+117) = 20 × 137 = 2740
an=a1+(n−1)d=dn+a1−ddn+a1−d=3n+7d=3a1=10a40=127
Σ40i=1ai=40⋅a1+a402=2740
nth term =3n+7
Hence first term (n=1)=3×1+7=10
40th term =3 ×40+7=127
Hence Sum of first 40 terms =(40/2)(10+127)
=2740
Sum of 40 terms = 600
Given:
The nth term of an AP is 4n-7
To find:
Sum of the first 40 terms
Solution:
Given nth term of AP = 4n - 7
⇒ First term, a₁ = 4(1) - 7 = 4 - 7 = - 3
⇒ Last term, a₄₀ = 4(40) - 7 = 40 - 7 = 33
As we know The formula for
Sum of n term in a AP, S
= n/2 (a₁+a)
Where a₁ = first term
a = nth or last term of AP
From above data sum of 40 terms in given AP
S₄₀ = 40/2 (-3+33) = 20(30) = 600
Sum of 40 terms = 600
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