Question

Find the sum of the following AP
-37,-33,-29,... to 12 terms

Answer 1

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• Find the sum of the following AP
• -37,-33,-29,... to 12 terms

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{!!ANSWER!!}}}}}}}$

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\large\mathcal{\bf{\boxed{\large\mathcal{~~-180~~}}}}}$

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

first term(a)=-37

numnber of terms(n)=12

difference=[(-33)-(-37)]=4

therefore the sum of 12 terms is

$= \frac{n(2a + (n - 1)d)}{2} \\ = \frac{12(2 ( - 37) + (12 - 1)4)}{2} \\ = 6( - 74 + 44) \\ = 6 \times( - 30) \\ = - 180$

$\huge\mathcal\green{\underline{hope\:\: this\:\: helps\:\: you}}$

Answer 2

Step-by-step explanation:

a=-37

d=t2-t1=-33+37=4

n=12

Sn=

$\frac{n}{2}$

2a+(n-1)d

S12=

$\frac{12}{2}$

2(-37)+11*4

=12/2 *-74+44

=6*(-30)

=-180