Question

The 5th and 12th term of an A.P. are 23 and 37 respectively. Find the sum of
the first 40 terms of the A.P.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\tt\it\bf\huge\it\bm{\mathcal{\fcolorbox{blue}{yellow}{\red{Given:-}}}}$

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Fifth term of AP (A5). :- 23

Twelve term of AP (A12) :- 37

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Sum of first 40 term of AP (S40):-

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$\tt\it\bf\huge\it\bm{\mathcal{\fcolorbox{black}{green}{\red{Solution:-}}}}$

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we know that :

$\tt{\mathcal{\blue{ An=a+(n-1)d}}}$

•°• A5 = a + (5-1)d

or,

➝ 23 = a +4d••••••••••••••(1)

Similarly,

➝ A12 = a + (12-1)d

➝ 37 = a + 11d•••••••••••••(2)

Using elimination method solve for (a) and (d) from equation (1) and (2):-

a+4d=23

a+11d=37

-_-___-__

➝ -7d = -14

➝ d = 14/7 = 2

putting the value of d in equation (1)

23=a+4d

➝ 23=a+4(2)

➝ 23 = a +8

➝ a=23-8

➝ a=15

•°• $\tt\huge{\red{ a= 15,d=2 }}$

$\tt{\mathcal{\blue{ Sn=n/2[2a+(n-1)d]}}}$

•°• S40 = 40/2{2(15) +(40-1)2}

➝20{30+39*2}

➝20{30 +78}

➝20{108}

➝2160

Hence, Sum of first 40 term= 2160

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