Question

If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P.
Also, find the sum of first 20 terms of this A.P.

Answer 1

e

S15= 15/2 ( 2* 2 + 14 * 5)

= 15/2 ( 4 + 70)

= 15/2 * 74

=15 * 37

=555

Answer 2

$\sf \green{Given :-}$

$\sf 8th \:term \:of \:A.P is\:37$

$\sf 15th\: term \: is \: 15 \: more\: than 12th\: term$

$\sf \green{To\: find:-}$

$\sf A.P \:series \: and \: sum\: of \: 20\: terms$

$\sf \green{Formulae\: used:-}$

$\sf\purple{ a_n = a+(n-1) d}$

$\sf\purple{ s_n = \dfrac{n}{2} [ 2a+(n-1)d]}$

$\sf \green{SOLUTION :-}$

$\sf As\: they\: given \: 8th\: term\: is 37$

$\sf So,$

$\sf a_n = a+(n-1) d$

$\sf n=8$

$\sf a_8 = a+(8-1) d$

$\sf {a_8 = a+7d}$

$\sf \purple{37 = a+7d } --- eq1$

$\sf Also,\: they\: given$

$\sf 15th\: term \: is \: 15 \: more\: than 12th\: term$

$\sf \purple{a_{15} = 15+a_{12}}$

$\sf a_{15} = a+(15-1) d$

$\sf \purple{a_{15} = a+14d }$

$\sf a_{12} = a+(12-1)d$

$\sf\purple{ a_{12} = a+11d}$

$\sf\purple{ a_{15} = 15+a_{12}}$

$\sf a+14d = 15+a+11d$

$\sf \not a+14d = 15+\not a+11d$

$\sf 14d = 15+11d$

$\sf 14d-11d =15$

$\sf 3d =15$

$\sf d= 15/3$

$\sf \purple{d=5}$

$\sf As \: we\: got\:" d" value\: substitute\: this\: value\: in \: eq1$

$\sf {37 = a+7(5) }$

$\sf {37 = a+35}$

$\sf 37-35 =a$

$\sf\purple {a= 2}$

$\sf Now\: we\: have \: "a" and\: "d" \: From\: this \: we\: can \: find \: the \: series \: of A.P$

$\sf a = first\: term$

$\sf d= common\: difference$

$\sf The\: series\: of\: A.P \: is$

$\sf\purple{ a, a+d , a+2d , a+3d . . . .}$

$\sf Substituting\: the\: values,$

$\sf 2, (2+5) ,[2+2(5)] , [2+3(5)] . . .$

$\sf 2, (7) ,(2+10) , (2+15). . .$

$\sf 2, 7,12 , 17. . .$

$\sf \purple{So,\:the \: series\: of \: A.P \:are\: 2,7, 12,17}$

$\sf Now, \: finding\: the\: sum \: of\: terms$

$\sf s_n = \dfrac{n}{2} [ 2a+(n-1)d]$

$\sf n= no. of\: terms \\ \sf a= first\: term \\ \sf d = common \: difference$

$\sf Here,$

$\sf n= 20 \\ \sf a= 2 \\ \sf d =5$

$\sf Substituting\: the\: values,$

$\sf s_{20}= \dfrac{20}{2} [ 2(2)+(20-1)5]$

$\sf s_{20}= 10 [ 4+(19)5]$

$\sf s_{20}= 10 [ 4+95]$

$\sf s_{20}= 10 (99)$

$\sf s_{20} = 990$

$\sf\purple{So, \: sum\: of\: 20\: terms\: in\: A.P\: is\: 990}$

----------------------------------

$\sf \green{Final\: answers :-}$

$\sf \purple{\:The \: series\: of \: A.P \:are\: 2,7, 12,17}$

$\sf\purple{The \: sum\: of\: 20\: terms\: in\: A.P\: is\: 990}$