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Question Number 112810 by Aina Samuel Temidayo last updated on 09/Sep/20

	Answered by Rasheed.Sindhi last updated on 10/Sep/20

	$t_{1}, t_{2}, t_{3}, . . ., t_{n}; n \in E . . . . . . . . . . . . (1)$ $S u m o f o d d n u m b e r e d t e r m s :$ $t_{1} + t_{3} + t_{5} + . . . + t_{n - 1} = 30 . . . . (2)$ $S u m o f e v e n n u m b e r e d t e r m s :$ $t_{2} + t_{4} + t_{6} + . . . + t_{n} = 24 . . . . . . . . (3)$ $L a s t t e r m e x c e e d s f i r s t b y 10.5 :$ $t_{n} - t_{1} = 10.5 . . . . . . . . . . . . . . . . . . . . . . (4)$ $n = ?$ $(1) : a, a + d, a + 2 d, . . . ., a + (n - 1) d$ $(4) : t_{n} - t_{1} = (a + (n - 1) d) - (a) = 10.5$ $d = \frac{10.5}{n - 1} = \frac{21}{2 (n - 1)} . . . . . ★$ $\Rightarrow n = \frac{10.5}{d} + 1$ $(2) : a + (a + 2 d) + (a + 4 d) + . . (a + \overset{―}{n - 2} d) = 30$ $N u m b e r o f t e r m s (N) = n / 2$ $F i r s t t e r m (A) = a$ $C o m m o n d i f f e r e n c e (D) = 2 d$ $S = \frac{N}{2} [2 A + (N - 1) . D]$ $S = \frac{n / 2}{2} [2 a + (\frac{n}{2} - 1) (2 d)] = 30$ $\frac{n}{4} [2 a + (n - 2) d] = 30$ $\frac{n a}{2} + \frac{n (n - 2) d}{4} = 30$ $\frac{n a}{2} = \frac{120 - n (n - 2) d}{4} . . . . . . . . . A$ $(3) : (a + d) + (a + 3 d) + (a + 5 d) + . . + a + (n - 1) d = 24$ $F i r s t t e r m (A) = a + d$ $C o m m o n d i f f e r e n c e (D) = 2 d$ $N u m b e r o f t e r m s (N) = n / 2$ $S = \frac{N}{2} [2 A + (N - 1) . D]$ $\frac{n / 2}{2} [2 (a + d) + (\frac{n}{2} - 1) (2 d)] = 24$ $\frac{n}{4} [2 a + n d] = 24$ $\frac{n a}{2} + \frac{n^{2} d}{4} = 24$ $\frac{n a}{2} = 24 - \frac{n^{2} d}{4}$ $\frac{n a}{2} = \frac{96 - n^{2} d}{4} . . . . . . . . . . . . . . . . . . B$ $F r o m A & B$ $\frac{120 - n (n - 2) d}{4} = \frac{96 - n^{2} d}{4}$ $120 - n (n - 2) d = 96 - n^{2} d$ $F r o m ★ : d = \frac{21}{2 (n - 1)}$ $120 - n (n - 2) (\frac{21}{2 (n - 1)}) = 96 - n^{2} (\frac{21}{2 (n - 1)})$ $\frac{240 (n - 1) - 21 n (n - 2)}{2 (n - 1)} = \frac{192 (n - 1) - 21 n^{2}}{2 (n - 1)}$ $240 (n - 1) - 21 n (n - 2) = 192 (n - 1) - 21 n^{2}$ $240 n - 240 - 21 n (n - 2) = 192 n - 192 - 21 n^{2}$ $48 n - 48 - 21 n (n - 2) = - 21 n^{2}$ $48 n - 48 \overset{\times}{- 21 n^{2}} + 42 n = \overset{\times}{- 21 n^{2}}$ $48 n - 48 + 42 n = 0$ $S o m e w h e r e t h e r e i s a n e r r o r . . .$
	Commented by Rasheed.Sindhi last updated on 10/Sep/20

	$I s t h e a n s w e r 8 ?$
	Commented by Rasheed.Sindhi last updated on 11/Sep/20
	$If we alter the question: sum of even numbered terms=30 sum of odd numbered terms=24 The answer will be 8. Otherwise check my above answer for errors because it gives fractional number as value of n.$
	$I f w e a l t e r t h e q u e s t i o n :$ $s u m o f e v e n n u m b e r e d t e r m s = 30$ $s u m o f o d d n u m b e r e d t e r m s = 24$ $T h e a n s w e r w i l l b e 8 .$ $O t h e r w i s e c h e c k m y a b o v e$ $a n s w e r f o r e r r o r s b e c a u s e i t$ $g i v e s f r a c t i o n a l n u m b e r a s v a l u e$ $o f n .$
	Commented by Aina Samuel Temidayo last updated on 10/Sep/20

	$I {don}^{'} t know the answer but 8 is part of the options I have here .$
	Answered by 1549442205PVT last updated on 11/Sep/20
	$Suppose the A.P has 2n terms and d is the its difference:{a_1 ,a_1 +d,...,a_1 +(2n−1)d} Since sum of odd numbered terms< Sum of even numbered terms,we infer a_1 is odd From the hypothesis a_(2n) −a_1 =10.5 we get (2n−1)d=10.5(1) Sum of n odd numbered terms equal to [a_1 +(2n−2)d]n/2=24(2) Sum of n even numbered terms equal to [a_2 +(2n−1)d].n/2=30(3) (2)(3)⇒((48)/(a_1 +(2n−2)d))=((60)/(a_2 +(2n−1)d)) ⇔48[a_2 +(2n−1)d]=60[a_1 +(2n−2)d] 4[a_1 +d+(2n−1)d]=5[a_1 +(2n−2)d] 4(a_1 +2nd)=5a_1 +10(n−1)d ⇔a_1 +(2n−10)d=0(4).From (1) we get d=((10.5)/(2n−1)).Replace into (4) we have a_1 +(2n−10)((10.5)/(2n−1))=0. ⇔(2n−1)a_1 +21n−105=0 ⇔a_1 =((105−21n)/(2n−1))⇔2a_1 =((210−42n)/(2n−1)) =−21+((189)/(2n−1))⇔(2n−1)(2a_1 +21)=189 ⇒(2a_1 +21)∣189=3^3 .7 ⇒(2a_1 +21)∈{1,3,7,9,21,27} determinant (((2a_1 +21),1,3,7,9,(21),(27)),((2n−1),(189),(63),(27),( 21),9,7),(a_1 ,(−10),(−9),(−7),( −6),0,3),(n,(95),(32),(14),(11),5,4)) The problem has six solutions : (a_1 ,2n)∈{(−10,190),(−9,64),(−7,28) ,(−6,22),(0,10),(3,8)}$
	$Suppose the A . P has 2 n terms and d$ $is the its difference : {a_{1}, a_{1} + d, . . ., a_{1} + (2 n - 1) d}$ $Since sum of odd numbered terms <$ $Sum of even numbered terms, we infer$ $a_{1} is odd$ $From the hypothesis a_{2 n} - a_{1} = 10.5$ $we get (2 n - 1) d = 10.5 (1)$ $Sum of n odd numbered terms equal to$ $[a_{1} + (2 n - 2) d] n / 2 = 24 (2)$ $Sum of n even numbered terms equal to$ $[a_{2} + (2 n - 1) d] . n / 2 = 30 (3)$ $(2) (3) \Rightarrow \frac{48}{a_{1} + (2 n - 2) d} = \frac{60}{a_{2} + (2 n - 1) d}$ $\Leftrightarrow 48 [a_{2} + (2 n - 1) d] = 60 [a_{1} + (2 n - 2) d]$ $4 [a_{1} + d + (2 n - 1) d] = 5 [a_{1} + (2 n - 2) d]$ $4 (a_{1} + 2 nd) = {5 a}_{1} + 10 (n - 1) d$ $\Leftrightarrow a_{1} + (2 n - 10) d = 0 (4) . From (1) we$ $get d = \frac{10.5}{2 n - 1} . Replace into (4) we have$ $a_{1} + (2 n - 10) \frac{10.5}{2 n - 1} = 0 .$ $\Leftrightarrow (2 n - 1) a_{1} + 21 n - 105 = 0$ $\Leftrightarrow a_{1} = \frac{105 - 21 n}{2 n - 1} \Leftrightarrow {2 a}_{1} = \frac{210 - 42 n}{2 n - 1}$ $= - 21 + \frac{189}{2 n - 1} \Leftrightarrow (2 n - 1) ({2 a}_{1} + 21) = 189$ $\Rightarrow ({2 a}_{1} + 21) ∣ 189 = 3^{3} . 7$ $\Rightarrow ({2 a}_{1} + 21) \in {1, 3, 7, 9, 21, 27}$ $\| \begin{matrix} {2 a}_{1} + 21 & 1 & 3 & 7 & 9 & 21 & 27 \\ 2 n - 1 & 189 & 63 & 27 & 21 & 9 & 7 \\ a_{1} & - 10 & - 9 & - 7 & - 6 & 0 & 3 \\ n & 95 & 32 & 14 & 11 & 5 & 4 \end{matrix} \|$ $The problem has six solution s :$ $(a_{1}, 2 n) \in {(- 10, 190), (- 9, 64), (- 7, 28)$ $, (- 6, 22), (0, 10), (3, 8)}$
	Commented by Aina Samuel Temidayo last updated on 10/Sep/20

	$I {don}^{'} t understand . We are to find the$ $number of terms . The answer to this$ $question is just a natural number .$
	Commented by 1549442205PVT last updated on 11/Sep/20
	$Number of terms is 2n.Since our problem has six answers corresponding to six first terms (a_1 ) ,so answer is a pair of (a_1 ,2n) Here,number of terms of six APs we find out are:2n∈{170,64,28,22,10,8} corresponding to first terms are a_1 ∈{−10,−9,−7,−6,0,3} Sometimes the answer of option question isn′t correct.Perhaps,you seen that thing on the forum.We must keep the bravery of mathematicians$
	$Number of terms is 2 n . Since our$ $problem has six answers$ $corresponding to six first terms (a_{1})$ $, so answer is a pair of (a_{1}, 2 n)$ $Here, number of terms of six APs we$ $find out are : 2 n \in {170, 64, 28, 22, 10, 8}$ $corresponding to first terms are$ $a_{1} \in {- 10, - 9, - 7, - 6, 0, 3}$ $Sometimes the answer of option$ $question {isn}^{'} t correct . Perhaps, you seen$ $that thing on the forum . We must keep$ $the bravery of mathematicians$
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