Question

The sum of first 10 terms of the A.P. : – 5,– 2, 1, … is ?


85
82
79
75
PLZ GUYS TELL ME THE CORRECT ANSWER ​

Answer 1

Answer:

85

Step-by-step explanation:

S10=10/2[2(-5)+(10-1)3] as Sn=n/2[2a+(n-1)d]

S10=5[-10+27]

S10=85

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Answer 2

Concept: A series of numbers is considered to be an arithmetic progression or sequence if there is a constant difference between the terms. The number sequence 5, 7, 9, 11, 13, 15,... is an example of an arithmetic progression with a common difference of 2. The following are some of the key phrases we  encounter in AP:

first term (a) ,

The sum of the first n terms $S_{n}$ ,

and the common difference (d)

Given: No. of terms = 10

           A.P. = -5, -2, 1, ......

To find: sum of first 10 terms

Solution: Here, in the A.P. the first term a = -5

               common difference d = -2 - (-5)

                                                    = -2 + 5

                                                    = 3

We first need to find the nth term i.e., 10th term of the AP using the formula

$a_{n} = a + (n-1)d$

where n =10

$a_{10} = -5 + (10 - 1)3\\a_{10} = -5 + (9)3\\a_{10} = -5 + 27\\a_{10} =22$

Now find the sum of first ten terms of an AP using the formula

$S = \frac{n}{2} [2a + (n - 1)d]$                 where, n = 10; a = -5; d = 3

$S = \frac{10}{2} [2*(-5) + (10 - 1)3]\\S = 5 [-10 + (9)3]\\S = 5 [ -10 + 27]\\S = 5 [ 17]\\S = 85$

Hence, the sum of first 10 terms of the AP -5, -2, 1, .... is 85.

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