What is the sum of the first ten terms of a_1 = -43, d=12 ?

3 Answers
TK02
Apr 30, 2018

S_10 = 110S10=110

Explanation:

a_1 = -43a1=−43
d = 12d=12
n = 10n=10

The formula for first 10 terms is:
S_n = 1/2n{2a +(n-1)d}Sn=12n{2a+(n−1)d}

S_10 = 1/2(10){2(-43) +(10-1)12}S10=12(10){2(−43)+(10−1)12}

S_10 = (5){-86 +(9)12}S10=(5){−86+(9)12}

S_10 = (5){-86 +108}S10=(5){−86+108}

S_10 = (5){22}S10=(5){22}

S_10 = 110S10=110

Karthik N
Apr 30, 2018

110
(Assuming the question refers to an Arithmetic Progression)

Explanation:

If I'm understanding this right(the lack of math notation makes it ambiguous!), this is an Arithmetic Progression with its first term a = -43a=−43 and common difference d = 12d=12.

The formula for the sum of the first nn terms of an A.P is S = [n(2a + (n-1)d)]/2S=n(2a+(n−1)d)2 .
Let's substitute a = -43a=−43, d = 12d=12 and n = 10n=10
S = [10(2(-43) + (10-1)12)]/2S=10(2(−43)+(10−1)12)2
S = 5(-86+ 9(12))S=5(−86+9(12))
S = 5(108 - 86) = 5(22)S=5(108−86)=5(22)

Thus the answer is 110.

Shwetank Mauria
Apr 30, 2018

Sum of first 1010 terms is 110110

Explanation:

Given first term of an arithmetic progression a_1a1 and common difference dd,

sum of first nnterms is given by

S_n=n/2(2a_1+(n-1)d)Sn=n2(2a1+(n−1)d)

Here a_1=-43a1=−43 and d=12d=12, hence

S_10=10/2(2xx(-43)+(10-1)*12)S10=102(2×(−43)+(10−1)⋅12)

= 5xx(-86+9xx12)5×(−86+9×12)

= 5xx(-86+108)5×(−86+108)

= 5xx225×22

= 110110

Related questions
Impact of this question
1711 views around the world
You can reuse this answer
Creative Commons License