Q-1-which-term-of-A-P-3-15-27-39-will-be-132-more-than-its-54th-term-

Question Number 7697 by Rohit 57 last updated on 09/Sep/16

Q . 1 w h i c h t e r m o f A . P : 3, 15, 27, 39 \dots

\dots w i l l b e 132 m o r e t h a n i t s 54 t h t e r m .

Commented by prakash jain last updated on 09/Sep/16

The given series

first term, a = 3

common difference d = 15 - 3 = 12

n^{t h} term = a + (n - 1) d

54^{t h} term = 3 + 53 \times 12 = 3 + 636 = 639

Value of required term = 639 + 132 = 771

771 = a + (n - 1) d = 3 + (n - 1) \times 12

12 (n - 1) = 768

n - 1 = 64

n = 65

65^{t h} term of the AP will be 132 more that

its 54^{th} term .

Commented by Rohit 57 last updated on 10/Sep/16

t h a n k u s i r

Commented by Rasheed Soomro last updated on 11/Sep/16

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A n o t h e r w a y

d = 15 - 3 = 12

132 = (\frac{132}{12}) \times 12 = 11 d

R e q u i e e d t e r m = 54 t h t e r m + 132 = 54 t h t e r m + 11 d

= (54 + 11) t h t e r m = 65 t h t e r m

p t h t r r m + q d = (p + q) t h t e r m

[a + (p - 1) d + q d = a + (\overset{―}{p + q} - 1) d]