The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A . P b e a, a + d, a + 2 d, a + 3 d, \dots, a + (n - 1) d \\ T a k i n g f i r s t a n d l a s t t e r m \\ a_{1} + a_{n} = a + a + (n - 1) d = 2 a + (n - 1) d \\ T a k i n g second a n d second l a s t t e r m \\ a_{2} + a_{n - 1} = (a + d) + [a + (n - 2) d] = 2 a + (n - 1) d = a_{1} + a_{n} \\ T a k i n g t h i r d f r o m t h e b e g i n n i n g a n d t h i r d f r o m t h e e n d \\ a_{3} + a_{n - 2} = (a + 2 d) + [a + (n - 3) d] = 2 a + (n - 1) d = a_{1} + a_{n} \\ F r o m t h e a b o v e p a t t e r n, w e o b s e r v e t h a t t h e s u m o f t e r m s equidistant f r o m t h e b e g i n n i n g \\ a n d t h e e n d i n a n A . P i s e q u a l t o t h e [f i r s t t e r m + l a s t t e r m] \\ H e n c e, t h e c o r r e c t v a l u e o f t h e f i l l e r i s [f i r s t t e r m + l a s t t e r m] . \end{array}$

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If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

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$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

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Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

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Alok Kumar Singh

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