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Answer 1

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If in a Arithmetic Progression, the common difference is 2 and the sum of first fifteen terms is 285. Then find the first term of the arithmetic progression??

First term of the A.P (a) = 5

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GIVEN -

Common Difference(d) = 2

no. of terms(n) = 15

Sum of First 15 terms $\tt( S_{15})$ = 285(S

15

TO FIND -

The first term of the arithmetic progression(a)

FORMULA USED -

$\tt\implies (S_{ {n}^{th} }) = \frac{n}{2}[2a + (n - 1)d]$

where,

n = no. of terms

a = first term of that A.P

d = common difference

CALCULATIONS

$\tt \: S _{ {n}^{th} } = sum \: of \: {n}^{th } term \: of \: the \: A.P$

$\tt\implies (S_{ {n}^{th} }) = \frac{n}{2}[2a + (n -$

Putting all the values,

$\tt\implies (S_{ {15}^{th} }) = \frac{15}{2}[2a + (15 - 1)2]$

$\tt\implies 285 = \frac{15}{2}[2a$

$\tt\implies \frac{285 \times 2}{15} = [2a$

$\tt\implies \frac{285 \times 2}{15} = (2a + 28)$

$\tt\implies {19 \times 2} = 2a + 28$

$\tt\implies 38 = 2a + 28$

$\tt\implies 38 - 28 = 2a$

$\tt \implies10 = 2a$

$\tt \implies5 = a$

Hence,

The first term of the arithmetic progression is 5.

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Answer 2

Answer:

GIVEN -

Common Difference(d) = 2

no. of terms(n) = 15

Sum of First 15 terms \tt( S_{15})(S

15

) = 285(S

15

TO FIND -

The first term of the arithmetic progression(a)

FORMULA USED -

\tt\implies (S_{ {n}^{th} }) = \frac{n}{2}[2a + (n - 1)d]⟹(S

n

th

)=

2

n

[2a+(n−1)d]

where,

n = no. of terms

a = first term of that A.P

d = common difference

CALCULATIONS

\tt \: S _{ {n}^{th} } = sum \: of \: {n}^{th } term \: of \: the \: A.PS

n

th

=sumofn

th

termoftheA.P

\tt\implies (S_{ {n}^{th} }) = \frac{n}{2}[2a + (n -⟹(S

n

th

)=

2

n

[2a+(n−

Putting all the values,

\tt\implies (S_{ {15}^{th} }) = \frac{15}{2}[2a + (15 - 1)2]⟹(S

15

th

)=

2

15

[2a+(15−1)2]

\tt\implies 285 = \frac{15}{2}[2a⟹285=

2

15

[2a

\tt\implies \frac{285 \times 2}{15} = [2a⟹

15

285×2

=[2a

\tt\implies \frac{285 \times 2}{15} = (2a + 28)⟹

15

285×2

=(2a+28)

\tt\implies {19 \times 2} = 2a + 28⟹19×2=2a+28

\tt\implies 38 = 2a + 28⟹38=2a+28

\tt\implies 38 - 28 = 2a⟹38−28=2a

\tt \implies10 = 2a⟹10=2a

\tt \implies5 = a⟹5=a

Hence,

The first term of the arithmetic progression is 5.