Question

If in an ap first term is 13 and common difference is -4 then sum of the first 10 term​

Answer 1

EXPLANATION.

In an A.P.

First term = a = 13.

Common difference = d = -4.

As we know that,

Sum of Nth term of an A.P.

⇒ Sₙ = n/2[2a + (n - 10d].

Using this formula in equation, we get.

⇒ S₁₀ = 10/2[2(13) + (10 - 1)(-4)].

⇒ S₁₀ = 5[26 + (9)(-4)].

⇒ S₁₀ = 5[26 - 36].

⇒ S₁₀ = 5[-10].

⇒ S₁₀ = -50.

                                                                                                                   

MORE INFORMATION.

Supposition of terms in A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Fourth terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Fifth terms as : a - 2d, a - d, a, a + d, a + 2d.

Answer 2

$\\ \large \underbrace {\underline\bold {Question :- }}$

If in an ap first term is 13 and common difference is -4 then sum of the first 10 term

$\\ \large \underbrace {\underline\bold {Solution :- }}$

Given that -

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: First \: term = a = 13$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: Common \: difference = d = (-4)$

Now , We have to find $\sf \: S_{10}$ ,

Using following formula ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{ \sf \: S_n = \: \frac{n}{2}[2a + (n - 1)d]}}$

Now, Simply put the values in formula,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: S_{10 }= \: \frac{10}{2}[2 \times 13 + (10 - 1)( - 4)]$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: S_{10 }= \: 5[26+ 9( - 4)]$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: S_{10 }= \: 5[26+ ( - 36)]$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: S_{10 }= \: 5[26 - 36]$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: S_{10 }= \: 5[ - 10]$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \underline {\boxed{ \orange{ \frak{\: S_{10 }= \: ( - 50)}}}}$

Hope it's helpful :)