Q-An-AP-has-even-number-of-terms-the-sum-of-odd-terms-is-24-and-of-even-terms-is-30-the-last-term-exceeds-the-first-term-by-10-1-2-find-AP-and-number-of-terms-

Question Number 7395 by 2402@gmail.com last updated on 26/Aug/16

Q . A n A P h a s e v e n n u m b e r o f t e r m s

t h e s u m o f o d d t e r m s i s 24 a n d

o f e v e n t e r m s i s 30 . t h e l a s t t e r m

e x c e e d s t h e f i r s t t e r m b y 10 \frac{1}{2} .

f i n d A P a n d n u m b e r o f t e r m s .

Commented by Yozzia last updated on 26/Aug/16

L e t n = 2 k b e n u m b e r o f t e r m s .

L e t u (t) = a + (t - 1) d f o r f i x e d a a n d d .

A l l e v e n t e r m s a r e u (2), u (4), u (6) \dots u (2 t), \dots u (2 k) .

∴ u (2 t) = a + (2 t - 1) d = h (t)

A l l o d d t e r m s a r e u (1), u (3), u (5), \dots, u (2 t - 1), \dots u (2 k - 1) .

∴ u (2 t - 1) = a + (2 t - 2) d = a + 2 (t - 1) d = g (t) .

N o w, \underset{t = 1}{\sum^{k}} g (t) = 24 a n d \underset{t = 1}{\sum^{k}} h (t) = 30 .

\underset{t = 1}{\sum^{k}} g (t) = 24 \Leftrightarrow \underset{t = 1}{\sum^{k}} (a + 2 d (t - 1)) = 24

\Leftrightarrow \underset{t = 1}{\sum^{k}} (a - 2 d) + 2 d \underset{t = 1}{\sum^{k}} t = (a - 2 d) k + d k (k + 1) = 24

o r 24 = k (a + d k - d) o r 24 = k (a + d k) - k d \dots . (1)

\underset{t = 1}{\sum^{k}} h (t) = 30 \Leftrightarrow \underset{t = 1}{\sum^{k}} (a + (2 t - 1) d) = 30

\Leftrightarrow \underset{t = 1}{\sum^{k}} (a - d) + 2 d \underset{t = 1}{\sum^{k}} t = 30 \Leftrightarrow k (a - d) + d k (k + 1) = 30

o r k (a + d k) = 30 \dots \dots (2)

C o m p a r i n g (1) a n d (2), (1) b e c o m e s

24 = 30 - k d \Rightarrow k d = 6 o r k = \frac{6}{d} \dots . (3)

T h e l a s t t e r m i s g i v e n b y u (2 k) = a + (2 k - 1) d

a n d u (2 k) - a = \frac{21}{2} . ∴ (2 k - 1) d = \frac{21}{2} .

B u t f r o m (3), k = \frac{6}{d} . ∴ d (\frac{12}{d} - 1) = \frac{21}{2}

12 - d = \frac{21}{2} \Rightarrow d = 12 - \frac{21}{2} = \frac{24 - 21}{2} = \frac{3}{2} .

∴ k = \frac{6}{3 / 2} = 4 .

I n (2) \Rightarrow 4 (a + 4 \times \frac{3}{2}) = 30

4 a + 24 = 30 o r a = \frac{3}{2} .

T h e A P i s h e n c e t h e s e q u e n c e {u (t)}_{t = 1}^{8}

w h e r e u (t) = \frac{3}{2} (1 + t - 1) = \frac{3 t}{2}, a n d t h e

n u m b e r o f t e r m s i s 2 k = 2 \times 4 = 8 .

Answered by Yozzia last updated on 27/Aug/16

C h e c k a n s w e r i n c o m m e n t s .

Answered by Rasheed Soomro last updated on 01/Sep/16

T e c h n i c a l l y S i m p l e A p p r o a c h

L e t a i s f i r s t t e r m a n d d i s a c o m m o n d i f f e r e n c e o f t h e A P .

T h e A P w i l l b e

a, a + d, a + 2 d, a + 3 d, a + 4 d, \dots a + 10 \frac{1}{2}

L e t t h e n u m b e r o f t e r m s i s 2 k

a + (2 k - 1) d = a + 10 \frac{1}{2} \Rightarrow (2 k - 1) d = 10 \frac{1}{2} \Rightarrow d = \frac{21}{2 (2 k - 1)}

N o w t h e A P w i l l b e

a, a + (\frac{21}{2 (2 k - 1)}), a + 2 (\frac{21}{2 (2 k - 1)}), a + 3 (\frac{21}{2 (2 k - 1)}), a + 4 (\frac{21}{2 (2 k - 1)}), \dots a + 10 \frac{1}{2}

T h e s u m o f o d d t e r m s

a + (a + 2 (\frac{21}{2 (2 k - 1)})) + (a + 4 (\frac{21}{2 (2 k - 1)})) + \dots + (a + 10 \frac{1}{2} - (\frac{21}{2 (2 k - 1)})) \dots . . k t e r m s

S = \frac{n}{2} (a + l) : a f i r s t t e r m, l l a s t t e r m, n n u m b e r o f t e r m s, S s u m .

S_{o} = \frac{k}{2} (a + (a - (\frac{21}{2 (2 k - 1)}) + 10 \frac{1}{2})) = 24

k (2 a + 10 \frac{1}{2} - (\frac{21}{2 (2 k - 1)})) = 48

2 a k + 10.5 k - k (\frac{21}{2 (2 k - 1)}) = 48 \dots \dots \dots \dots \dots . (i)

S u m o f e v e n t e r m s

(a + (\frac{21}{2 (2 k - 1)})) + (a + 3 (\frac{21}{2 (2 k - 1)})) + (a + 5 (\frac{21}{2 (2 k - 1)})) + \dots a + 10 \frac{1}{2} \dots . k t e r m s

S_{e} = \frac{k}{2} ((a + \frac{21}{2 (2 k - 1)}) + (a + 10 \frac{1}{2})) = 30

k (2 a + 10 \frac{1}{2} + (\frac{21}{2 (2 k - 1)})) = 60

2 a k + 10.5 k + k (\frac{21}{2 (2 k - 1)}) = 60 \dots \dots \dots \dots \dots . (i i)

(i i) - (i) : 2 k (\frac{21}{2 (2 k - 1)}) = 12 \Rightarrow k (\frac{21}{2 (2 k - 1)}) = 6 \Rightarrow 21 k = 12 (2 k - 1)

24 k - 21 k = 12 \Rightarrow k = 4

(i i) + (i) : k (4 a + 21) = 108 \Rightarrow 4 a + 21 = 108 / k = 108 / 4 = 27

4 a = 6 \Rightarrow a = \frac{3}{2}

T h e f i r s t t e r m o f t h e A P i s \frac{3}{2} a n d

c o m m o n d i f f e r e n c e i s \frac{21}{2 (2 k - 1)} = \frac{21}{2 (2 (4) - 1)} = \frac{3}{2}

H e n c e T h e A P i s

\frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12 (8 t e r m s)