Question

The first term of an A.P is -3 and the 10th term is 15, then S10 = --

Answer 1

Given :

• First term of AP= -3

• 10th term = 15

To find :

• Sum of first 10 terms ( S10)

Formula used :

• Sn = (n/2)(a+l)

where :-

• a = first term
• n = number of terms
• l = last term
• Sn= sum of n terms

Solution :

First term (a) = -3

Last term (l) = 15

⟹ S10 = (10/2)(-3+15)

⟹ S10 = 5 (12)

⟹ S10 = 5×12

⟹ S10 = 60

Therefore sum of first 10 terms = 60

Answer : 60

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$\large\bf\red{Additional-Information}$

1. Tn = a+(n-1)d

2. Sn = n/2[2a+(n-1)d]

3. Sn = n/2(a+l)

where :-

• a = first term
• n = number of terms
• d = common difference
• Tn = nth term

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Answer 2

Given ,

First term (a) = -3

10th term $\tt( a_{10})$ = 15

We know that , the sum of n terms of an AP is given by

$\boxed{ \tt{S_{n} = \frac{n}{2} \{a + a_{n}\}}}$

Thus ,

$\tt S_{10} = \frac{10}{2} \{ - 3 + 15\}$

$\tt S_{10} = 5 \times 12$

$\tt S_{10} =60$

Therefore , the sum of first ten terms of AP is 60

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