Question

determine the sum of the first 20 terms of the arithmetic series in which the 15th term si 107 and the terms decrease by 3. plz help, i've got a test tmrw. i need to see the procedure

Answer 1

you mean to say that 14th term is 104 or 16th term is 104?

Answer 2

i think the 16th term is 104

Answer 3

14th is 110

Answer 4

ok.considering that increasing terms decrease by 3
so, we have
\[a_{15}=107\]
for nth term in an arihthmetic progression,
\[a _{n}=a+(n-1)d\]
where a is the first term and d is the common difference.
\[a _{15}=a+(15-1)(-3)\]
107=a-42
a=149

Answer 5

similarly as we have to determine the sum of first 20 terms
\[a _{20}=a+(20-1)(-3)\]
so
\[a _{20}=149-57=92\]
now to find sum of terms
\[S _{n}=\frac{ n }{ 2 }[a+l]\]
where n is the number of terms, a is first term and l is last term.

Answer 6

\[S _{20}=\frac{ 20 }{ 2 }[149+92]\]
=10*241=2410

Answer 7

alternatively , you can use this formula
\[S _{n}=\frac{ n }{ 2 }[2a+(n-1)d]\]
where n is the number of terms , a is first term and d is common difference.

Answer 8

we already solved for a in the first part, so why did we solve for a again?

Answer 9

\[S _{20}=\frac{ 20 }{ 2 }[2*149+(20-1)(-3)]=2410\]

Answer 10

first you got 149 n then 92

Answer 11

in above formula we have to find first and last terms to find the sum.
so later we found 20th term of the series which is 92

Answer 12

ohh...i see

Answer 13

i get it, thank you!

Answer 14

you are welcome.. @qwertykeyboard