Question

(iii) The first term of an A.P. is -5 and the 6th term is 45. Find S6.​

Answer 1

Solution

Given :-

• First term of an A.P. = -5
• Sixth term of an A.P. = 45

Find :-

• Sum of sixth terms of A.P. series

Explanation

Using Formula

$\red{\dag\boxed{\underline{\tt{\green{\:Sum\:of\:n^{th}\:term\:=\:\dfrac{n(A+L)}{2}}}}}}$

Where,

• n = 6
• A = -5
• L = 45

Keep all above Values.

➠ sum of sixth terms = 6(-5 + 45)/2

➠ sum of sixth terms = 6(40)/2

➠ sum of sixth terms = 3 × 40

➠ sum of sixth terms = 120

Hence

• Sum of sixth terms will be = 120

___________________

Note

• If given first term & common defference then we use this formula for sum .

$\red{\dag\boxed{\underline{\tt{\orange{\:S_{n}\:=\:\dfrac{n}{2}\times\big(a+(n-1)d\big)}}}}}$

Where,

• a = first term
• n = total number of terms
• d = common Defference

__________________

Answer 2

Step-by-step explanation:

Solution

Given :-

First term of an A.P. = -5

Sixth term of an A.P. = 45

Find :-

Sum of sixth terms of A.P. series

Explanation

Using Formula

\red{\dag\boxed{\underline{\tt{\green{\:Sum\:of\:n^{th}\:term\:=\:\dfrac{n(A+L)}{2}}}}}}†

Sumofn

th

term=

2

n(A+L)

Where,

n = 6

A = -5

L = 45

Keep all above Values.

➠ sum of sixth terms = 6(-5 + 45)/2

➠ sum of sixth terms = 6(40)/2

➠ sum of sixth terms = 3 × 40

➠ sum of sixth terms = 120

Hence

Sum of sixth terms will be = 120

___________________

Note

If given first term & common defference then we use this formula for sum .

\red{\dag\boxed{\underline{\tt{\orange{\:S_{n}\:=\:\dfrac{n}{2}\times\big(a+(n-1)d\big)}}}}}†

S

n

=

2

n

×(a+(n−1)d)

Where,

a = first term

n = total number of terms

d = common Defference

__________________