Answer:
1540
Step-by-step explanation:
This is an arithmetic progression.
a = first term = 10
Common difference = d = second term - first term
= 12 - 10
d = 2
Last term = l = 78
First we have to find how many terms are there in the sequence using the formula: l = a + (n-1)*d
78 = 10 + (n -1) * 2
78 -10 = (n -1)*2
68 = (n -1) *2
68 ÷2 = n -1
34 = n - 1
34 + 1 = n
n = 35
There are 35 terms.
[tex]\sf \boxed{\test{\bf Sum = $\dfrac{n}{2}(a +l)$}}[/tex][tex]\sf \boxed{\text{\bf Sum =$\dfrac{n}{2}(a+l) $}}[/tex]
[tex]\sf =\dfrac{35}{2}(10+78)\\\\ =\dfrac{35}{2}*88\\\\ = 35 * 44\\\\= 1540[/tex]
Step-by-step explanation:
This is an arithmetic progression.
a = first term = 10
Common difference = d = second term - first term
= 12 - 10
d = 2
Last term = l = 78
First we have to find how many terms are there in the sequence using the formula: l = a + (n-1)*d
78 = 10 + (n -1) * 2
78 -10 = (n -1)*2
68 = (n -1) *2
68 ÷2 = n -1
34 = n - 1
34 + 1 = n
n = 35
There are 35 terms.
\sf \boxed{\test{\bf Sum = $\dfrac{n}{2}(a +l)$}} \sf \boxed{\text{\bf Sum =$\dfrac{n}{2}(a+l) $}}
Sum =
2
n
(a+l)
\begin{gathered}\sf =\dfrac{35}{2}(10+78)\\\\ =\dfrac{35}{2}*88\\\\ = 35 * 44\\\\= 1540\end{gathered}
=
2
35
(10+78)
=
2
35
∗88
=35∗44
=1540
