Question

find the sum (-5) + (-8) + (-11) + ...........+ (-62)​

Answer 1

Step-by-step explanation:

Given AP:

2, 5, 8, 11.... 62

Here,

a = 2, d = 8 - 5 = 3, an = 62

Here,

We aren't provided with the number of terms of which we have to find the sum of.

But we are given with the last term of the term, by which we are surely find out the number of terms.

By using general term,

an = a + (n-1)d

62 = 2 + (n - 1)3

62 = 2 + 3n - 3

62 = - 1 + 3n

63 = 3n

n = 21

Now, we are having number of terms.

So, by using sum of first n terms,

S_n = \frac{n}{2}[2a+(n-1)d]Sn=2n[2a+(n−1)d]

S_{21} = \frac{21}{2}[2(2) + (21-1)3]S21=221[2(2)+(21−1)3]

S_{21} = \frac{21}{2}[4 + 60]S21=221[4+60]

S_{21} = \frac{21}{2}(64)S21=221(64)

S_{21} = \frac{21}{\cancel{2}}(\cancel{64})S21=221(64)

S_{21} = 21 \times 32S21=21×32

S21 = 672

mark me as brainlist

Answer 2

$\large\underline{\sf{Solution-}}$

Given AP, -5, -8, -11, ..., -62

Here, first term (a) = -5, common difference (d) = {(-8) - (-5)} = {-8 + 5} = -3 and last term (l) = -62.

Let the total number of terms be n. Then,

$\\ \mathrm{T_{n} = -62 \implies a + (n -1)d = -62.} \\ \\\mathrm{\implies \: - 5 + (n - 1) - 3 = - 62} \\ \\ \mathrm{\implies \: - 5 - 3n + 3 = - 62} \\ \\ \mathrm{\implies - 3n - 2 = - 62} \\ \\ \mathrm{\implies \: - 3n = - 60} \\ \\ \mathrm{\implies \: n = \dfrac{ - 60}{ - 3} } \\ \\ \mathrm{\therefore \: n = 20}$

Required Sum = $\sf{\dfrac{n} {2} × (a + l)}$

$\mathrm{ = \dfrac{20}{2}( - 5 + - 62) } \\ \\ \mathrm{ \dfrac{20}{2} \times - 67 } \\ \\ \mathrm{10 \times - 67} \\ \\ \mathrm{ - 670}$

Hence, the sum of the AP is -670.

$\rule{200pt} {2pt}$