Question

the sum of the series 9, 5, 1...... to 100 terms is​

Answer 1

Given that the first term of series = 9

Second term = 5 and third term = 1

We can see that, 5 - 9 = 1 - 5

Hence, given series is an A.P.

Here, common difference = 5 - 9

→ d = - 4

and, a = 9

Number of terms (n) = 100

Now, we know that sum of an A.P. =

$\frac{n}{2} \{2a + (n - 1)d \}$

So, insert the values of a, n and d to get the sum.

$\sf\frac{100}{2}\{2(9) + (100 - 1)(-4)\}$

$\implies\sf 50\{18 + (-396)\}$

$\implies\sf 50(-378)$

$\implies\sf -18900$

$\huge\mathfrak{Answer}$

$\sf Sum \ of \ series \ is \ -18900$

Answer 2

$\large{\color{blue}{\boxed{\boxed{-18900}}}}$

Here,

first term,a= 9

common difference,d= a2 - a1 = 5-9 = -4

sum of this A.P. upto 100 terms,

S100= $\frac{n}{2}$(2a+(n-1)d)

S100= $\frac{100}{2}$(2×9 + (100-1)-4)

S100= 50(18 + (99)-4)

S100= 50(18 - 396)

S100= 50 × (-378)

S100= -18900

Hope it helps!!!☺

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