Question

Find the sum of all terms of an A.P. if the common difference is -2, its first term
is 100 and the last term is - 10.

Answer 1

Answer:

2530

Step-by-step explanation:

a=100, d=-2,

Last term Tn =-10

Tn=a+(n-1)d

-10=100+(n-1)(-2)

-110=(n-1)(-2)

n-1=-110/-2=55

Now Sum of all terms

=55/2( 200 + 54*-2 )

=55/2*92

=55*46=2530

Answer 2

$\sf Given \begin{cases} & \sf{First\:term,\: a = \bf{100}} \\ & \sf{Common\:diffrence,\:d = \bf{-2}} \\ & \sf{Last\:term,\: l = \bf{-10}} \end{cases}$

To find: Sum of all terms of an A.P.

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

☯ Let number of terms of an A.P. be "n".

⠀⠀⠀⠀

$\bf{\dag}\;{\underline{\frak{n^{th}\:term\:of\:an\:A.P.\:is\:given\;by,}}}$

$\star\;{\boxed{\sf{\pink{a_n = a + (n - 1)d}}}}$

$:\implies\sf - 10 = 100 + (n - 1) \times - 2$

$:\implies\sf - 10 - 100 = (n - 1) \times - 2$

$:\implies\sf - 110 = (n - 1) \times - 2$

$:\implies\sf (n - 1) = \cancel{ \dfrac{-110}{-2}}$

$:\implies\sf (n - 1) = 55$

$:\implies{\underline{\boxed{\frak{\purple{n = 54}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Number\:of\:terms\:in\:an\:A.P.\:is\: {\textsf{\textbf{54}}}.}}}$

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$\bf{\dag}\;{\underline{\frak{Sum\:of\:all\:terms\:of\:A.P.\:is\:given\;by,}}}$

$\star\;{\boxed{\sf{\pink{S_n = \dfrac{n}{2}(a + l)}}}}$

$:\implies\sf S_{54} = \dfrac{54}{2}(100 + (-10))$

$:\implies\sf S_{54} = \dfrac{54}{2}(100 - 10)$

$:\implies\sf S_{54} = \dfrac{10}{ \cancel{2}} \times \cancel{90}$

$:\implies\sf S_{54} = 54 \times 45$

$:\implies{\underline{\boxed{\frak{\purple{S_{54} = 2430}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Sum\:of\:all\:terms\:of\:an\:A.P.\:is\: {\textsf{\textbf{2430}}}.}}}$