8)There are 20 terms in an arithmetic sequence. Sum of 1st and last term is 90. a)What is the sum of second and 19th term? b)Find the sum of first 20 terms.
Answers
Step-by-step explanation:
ANSWER:
The sum of 20 terms = - 730
GIVEN:
First term of arithmetic series = - 122.
Number of terms = 20.
Common difference = 9.
TO FIND:
The sum of 20 terms.
EXPLANATION:
\boxed{\bold{\large{\blue{S_n = \frac{n}{2}\{2a + (n-1) d\}}}}}
S
n
=
2
n
{2a+(n−1)d}
n = 20
a = - 122
d = 9
\sf \dashrightarrow S_{20} = \dfrac{20}{2}\{2( - 122) + (20-1) 9\}⇢S
20
=
2
20
{2(−122)+(20−1)9}
\sf \dashrightarrow S_{20} = 10\{- 244 + (19) 9\}⇢S
20
=10{−244+(19)9}
\sf \dashrightarrow S_{20}= 10\{- 244 + 171\}⇢S
20
=10{−244+171}
\sf \dashrightarrow S_{20} = 10(- 73)⇢S
20
=10(−73)
\sf \dashrightarrow S_{20} = - 730⇢S
20
=−730
\orange{\large{\bold{\checkmark}}\boxed{ \bold{ \large{ \gray{Sum \ of \ first\ 20 \ terms = -730}}}}}✓
Sum of first 20 terms=−730
VERIFICATION:
\boxed{\bold{\large{\red{n = \dfrac{l-a}{d} +1}}}}
n=
d
l−a
+1
n = 20
d = 9
a = - 122
\sf \leadsto 20 = \dfrac{l + 122}{9} +1⇝20=
9
l+122
+1
\sf \leadsto 20 = \dfrac{l + 122 + 9}{9}⇝20=
9
l+122+9
\sf \leadsto 180 =l + 131⇝180=l+131
\sf \leadsto l = 49⇝l=49
\boxed{\bold{\large{\purple{S_n = \frac{n}{2}(a + l)}}}}
S
n
=
2
n
(a+l)
n = 20
a = - 122
l = 49
\sf \leadsto S_{20} = \dfrac{20}{2}( - 122 + 49)⇝S
20
=
2
20
(−122+49)
\sf \leadsto S_{20} = 10( -73)⇝S
20
=10(−73)
\sf \leadsto S_{20} = - 730⇝S
20
=−730
HENCE VERIFIED.
Answer:
Step-by-step explanation:
8 To find the common difference of an arithmetic sequence, use the basic formula is an=a1+(n-1)d
an: the nth term in the sequence (90)
a1: the first term in the sequence (-5)
n :the number of terms in the sequence (20)
d :the common difference
First, you subtract the last term by the first term.
Next, divide the difference by the value of the “number of terms minus 1” to isolate the “d”, and you will have your answer.
For your question, the first step should be 90 minus -5, which equals 95, and the second step 95 divided by 19, which equals 5.