Question

Find the sum.
10+12+14+...+78

Answer 1

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

Answer 2

1540

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 = $(n)/(2)(a +l)$}}`
`\sf \boxed{\text{\bf Sum =$(n)/(2)(a+l) $}}`

`\sf =(35)/(2)(10+78)\\\\ =(35)/(2)*88\\\\ = 35 * 44\\\\= 1540`