Question

Find the sum of first 22 terms of an A.P. in which d=22 and a 22 =149

Answer 1

Sum of an AP is (1/2)(number of terms) (first term + last term)

=(1/2)(n) [a + a+(n-1)d]

Here,

d=22

n=22

(a22)=149=(a1)+(21*d)

149=(a1)+(21*22)

(a1) =149-462=-313

Sum of AP=(0.5*22)(-313+149)

=(0.5*22)(-164)

=-1804

Sum of the above AP is -1804.

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Answer 2

Answer:

$\huge{\pink{\boxed{\green{\sf{S_{22}=-1804}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{[SOLUTION-]}}}}}}}$

$\: \: { \orange{given}} \\ { \pink{ \boxed{ \green{d = 22}}}} \\ { \pink{ \boxed{ \green{ a_{22} = 149}}}} \\ { \pink{ \boxed{ \green{ n = 22}}}} \\ \\ { \blue{to \: find}} \\ { \purple{ \boxed{ \red{ s_{22} =? }}}}$

☆ According to given question:

$\to a_{22} =149 \\ \to a + 21d = 149 \\ \to a + 21 \times 22 = 149 \\ \to a + 462 = 149 \\ \to a = 149 - 462 \\ { \boxed{\to a = - 313}}$

☆ We know all values that required to find sum:

☆ So put these values in basic formula:

$\to s_{n} = \frac{n}{2} (a + a_{n}) \\ \to s_{n} = \frac{22}{2} ( - 313 + 149) \\ \to s_{n} =11( - 164) \\ { \pink{ \boxed{ \green{\therefore s_{n} = - 1804}}}}$

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