Question

Find the sum of first 30 terms of arithmetic progression whose third term is 11 and 5th term is 17​

Answer 1

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✦Required Answer:

♦️ GiveN:

• Third term of an AP = 11
• Fifth term of an AP = 17

♦️ To FinD:

• Sum of first 30 terms of the AP

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✦ How to Solve?

As the question of AP (Arithmetic Progression) concept. We just need to know certain formulas like nth term of AP and sum of n terms. The arithmetic progression has a common difference between their terms.

$\large{\underline{\underline{\purple{\rm{nth\: term \:of\: an \:AP :}}}}}$

$\large{ \boxed{ \rm{T_n = a + (n - 1)d}}}$

Where, Tn = nth term, a = first term, n = no. of terms and d = common difference.

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$\large{\underline{\underline{\purple{\rm{Sum\:of \:n\: terms \:of\: an\: AP :}}}}}$

$\large{ \boxed{ \rm{S_n = \frac{n}{2} \bigg(\: 2a + (n - 1)d} \bigg)}}$

Where, Sn = Sum of n terms, rest are same.....

So, let's solve this question, by using these formulae.

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✦ Solution:

We have,

$\large{ \rm{ \longrightarrow \: a_3 = 11}} \\ \\ \rm{ \dag{ \red{by \: using \: nth \: term \: formula....}}} \\ \\ \large{ \rm{ \longrightarrow \: a + 2d = 11........(1)}}$

Similarly,

$\large{ \rm{ \longrightarrow \: a_5 = 17}} \\ \\ \rm{ \dag{ \red{by \: using \: nth \: term \: of \: formula}}} \\ \\ \large{ \rm{ \longrightarrow \: a + 4d = 17........(2)}}$

Subtracting eq.(1) from eq.(2),

$\large{ \rm{ \longrightarrow \: a + 4d - (a + 2d) = 17 - 11}} \\ \\ \large{ \rm{ \longrightarrow \: \cancel{ a} + 4d - \cancel{a} - 2d = 6}} \\ \\ \large{ \rm{ \longrightarrow \: 2d = 6}} \\ \\ \large{ \rm{ \longrightarrow \: d = 3}}$

Putting value of d in eq.(1),

$\large{ \rm{ \longrightarrow \: a + 2(3) = 11}} \\ \\ \large{ \rm{ \longrightarrow \: a + 6 = 11}} \\ \\ \large{ \rm{ \longrightarrow \: a = 5}}$

So, we have our 1st term = 5 and c.d = 3, let's find the sum of 30 terms by using sum of n terms formula,

By using formula,

$\large{ \rm{ \longrightarrow \: S_{30} = \frac{30}{2} \bigg(2(5) + (30 - 1)3 \bigg)}} \\ \\ \large{ \rm{ \longrightarrow \: S_{30} = 15 \bigg((10 + 87) \bigg)}} \\ \\ \large{ \rm{ \longrightarrow \: S_{30} = 15 \times 97}} \\ \\ \large{ \rm{ \longrightarrow \: S_{30} = \boxed{ \rm{ \red{1455}}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \rm{ \green{hence \: soved \: \dag}}}}}}$

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Answer 2

$\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}$

$\rm{Find \ the \ sum \ of \ first \ 30 \ terms \ of }$

$\rm{arithmetic \ progression \ whose \ 3^{rd} \ term}$

$\rm{is \ 11 \ and \ 5^{th} \ term \ is \ 17}$

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$\purple{\underline{\underline{\orange{\boxed{\boxed{\purple{\star{\sf Answer:}}}}}}}}$

$\rm\green{The \ sum \ of \ first \ 30 \ terms \ of \ the \ AP}$

$\rm{\red{\boxed{\boxed{\green{\star{S_{30} = 1455}}}}}}$

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$\sf\large\underline\pink{Given:}$

$\rm{The \ 3^{rd} \ term \ of \ the \ AP}$

$\bf\implies{t(3) \ = \ 11}$

$\rm{The \ 5^{th} \ term \ of \ the \ AP}$

$\bf\implies{t(5) \ = \ 17}$

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$\sf\large\underline\blue{To \ Find}$

$\rm{The \ sum \ of \ the \ first \ 30 \ terms \ of \ the \ AP}$

$\rm{i.e. \ S(30) \ = \ ?}$

_________________________________________

$\green{\underline{\underline{\red{\boxed{\boxed{\purple{\star{\sf Solution:}}}}}}}}$

$\rm{We \ are \ given \ that}$

$\rm{The \ 3^{rd} \ term \ of \ the \ AP}$

$\rm\implies{t(3) \ = \ 11}$

$\rm{The \ 5^{th} \ term \ of \ the \ AP}$

$\rm\implies{t(5) \ = \ 17}$

$\rm{We \ know}$

$\rm{The \ n^{th} \ term \ of \ AP}$

$\rm{\green{\boxed{\orange{\star{T(n) = a + (n - 1)d}}}}}$

$\rm{Where, \ T(n) \ = \ n^{th} \ term,}$

$\rm{a \ = \ first \ term ,}$

$\rm{n \ = \ number \ of \ terms }$

$\rm{d \ = \ common \ difference }$

$\rm\therefore{According \ to \ first \ condition ,}$

$\rm\implies{t(3) \ = \ 11}$

$\rm\therefore{t(3) \ = \ 2a \ + \ (3-1)d}$

$\rm\therefore{11 \ = \ a \ + \ 2d}$

$\rm\therefore{a \ + \ 2d \ = \ 11 \ ..................(a)}$

$\rm{Now,}$

$\rm{According \ to \ second \ condition}$

$\rm\implies{t(5) \ = \ 17}$

$\rm\therefore{t(5) \ = \ a \ + \ (5-1)d}$

$\rm\therefore{17 \ = \ a \ + \ 4d}$

$\rm\therefore{ a \ + \ 4d \ = \ 17 \ ...................(b)}$

$\rm{Subtracting \ equation \ () \ from \ (b) }$

$\rm{a \ + \ 4d \ = \ 17 }$

$\rm{-}$

$\rm{a \ + \ 2d \ = \ 11 }$

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$\rm{2d \ = \ 6}$

$\rm{d \ = \ \frac{6}{2}}$

$\rm{\blue{\boxed{\pink{\therefore{d \ = \ 3}}}}}$

$\rm{Now, \ putting \ d=3 \ in \ equation \ (a)}$

$\rm{We \ get,}$

$\rm{a \ + \ 6 \ = \ 11}$

$\rm{a \ = \ 11 \ - \ 6}$

$\rm{\blue{\boxed{\pink{\therefore{a \ = \ 5}}}}}$

$\rm{Now,}$

$\rm{Sum\:of \:n\: terms \:of\: an\: AP :}$

$\rm{\green{\boxed{\orange{\star{S(n) = \frac{n}{2} ( 2a + (n - 1)d)}}}}}$

$\rm{Where, \ Sn \ = \ Sum \ of \ n \ terms}$

$\rm{We \ have \ to \ find \ sum \ of \ first \ 30 \ terms \ of \ AP}$

$\rm\therefore{By \ using \ formula, }$

$\rm{ \implies {S_{30} = \frac{30}{2} (2(5) + (30 - 1)3 )}}$

$\rm{ \implies{S_{30} = 15 (10 + 87) }}$

$\rm{ \implies{S_{30} = 15 \times 97}}$

$\rm{\red{\boxed{\boxed{\green{\star{S_{30} = 1455}}}}}}$

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$\rm\red{Hope \ it \ helps \ :))}$