Question

Find the sum of the series 1+4+7+10+...+118

Answer 1

Answer:

The given sequence is an Arithmetic Progression.

First term (a) = 1

Common difference(d) = 4 - 1 =3

Last term (l) = 118

Number of terms in series = n

Therefore,

l = a + (n-1)d

118 = 1 +(n-1)×3

117 ÷ 3 = n - 1

39 +1 = n

n = 40

Now use the formula

Sum = n/2 [ a + l]

=40/2 [ 1 + 118]

= 20×119

= 2380

Answer 2

Given :-

AP = 1, 4, 7, 10 .... 118

here, we've to find the sum of the series.

first term = a = 1

first term = a = 1common difference = d = 4 - 1 = 3

first term = a = 1common difference = d = 4 - 1 = 3An = 118

no. of terms isn't given so first of all, we've to find n

general form of nth term = a + (n - 1)d

➡ a + (n - 1)d = 118

➡ 1 + (n - 1)3 = 118

➡ 1 + 3n - 3 = 118

➡ 3n = 118 + 3 - 1

➡ 3n = 120

➡ n = 120/3

➡ n = 40

thus there are 40 terms.

now, their sum = n/2 (a + l) (l = last term)

= 40/2 (1 + 118)

= 20 × 119

= 2380

hence, the sum of the given series is 2380