When the terms of a Geometric Progression(GP) with common ratio r=2 is added to the corresponding terms of an Arithmetic Provression (AP), a new sequence is formed. If the first terms of the GP and AP are the same and the first three terms of the new sequence are 3, 7 and 11 respectively, find the n^(th) term of the seauence.


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Question Number 111342 by pete last updated on 03/Sep/20

$When the terms of a Geometric Progression (GP)$ $with common ratio r = 2 is added to the corresponding terms$ $of an Arithmetic Provression (AP),$ $a new sequence is formed . If the first terms$ $of the GP and AP are the same and the first$ $three terms of the new sequence are$ $3, 7 and 11 respectively, find the n^{th} term$ $of the seauence .$
	Answered by Rasheed.Sindhi last updated on 03/Sep/20
	$General term of new sequence T_n =ar^(n−1) +a+(n−1)d =a{(2)^(n−1) +1}+(n−1)d T_1 =a{(2)^(1−1) +1}+(1−1)d=3 =2a=3⇒a=3/2 T_2 =a{(2)^(2−1) +1}+(2−1)d=7 =(3/2)(3)+d=7 ⇒d=7−9/2=5/2 T_n =((3/2))(2^(n−1) +1)+(n−1)((5/2)) T_n =((3.2^(n−1) +5n−2)/2) T_1 =((3.2^0 +3)/2)=3 T_2 =((3.2+8)/2)=7 T_3 =((3.2^(3−1) +5(3)−2)/2)=((12+13)/2)=((25)/2) ≠11 ⊚ ′More than sufficient′ data ⊡Inconsistent data$
	$G e n e r a l t e r m o f n e w s e q u e n c e$ $T_{n} = {a r}^{n - 1} + a + (n - 1) d$ $= a {{(2)}^{n - 1} + 1} + (n - 1) d$ $T_{1} = a {{(2)}^{1 - 1} + 1} + (1 - 1) d = 3$ $= 2 a = 3 \Rightarrow a = 3 / 2$ $T_{2} = a {{(2)}^{2 - 1} + 1} + (2 - 1) d = 7$ $= (3 / 2) (3) + d = 7$ $\Rightarrow d = 7 - 9 / 2 = 5 / 2$ $T_{n} = (\frac{3}{2}) (2^{n - 1} + 1) + (n - 1) (\frac{5}{2})$ $T_{n} = \frac{3 . 2^{n - 1} + 5 n - 2}{2}$ $T_{1} = \frac{3 . 2^{0} + 3}{2} = 3$ $T_{2} = \frac{3.2 + 8}{2} = 7$ $T_{3} = \frac{3 . 2^{3 - 1} + 5 (3) - 2}{2} = \frac{12 + 13}{2} = \frac{25}{2}$ $\neq 11$ $⊚^{'} M o r e t h a n {s u f f i c i e n t}^{'} d a t a$ $⊡ I n c o n s i s t e n t d a t a$
	Commented by Aina Samuel Temidayo last updated on 03/Sep/20

	$Please what do you mean by$ $inconsistent data ?$
	Commented by Rasheed.Sindhi last updated on 03/Sep/20

	$C o n t r d i c t i o n b e t w e e n d a t a . L i k e$ $h e r e T_{3} = 11 i s g i v e n b u t u s i n g$ $o t h e r d a t a w e c a l c u l a t e T_{3} = 25 / 2$ $T_{3} s a m e t i m e {c a n}^{'} t b e e q u a l t o$ $t w o v a l u e s!$
	Commented by Aina Samuel Temidayo last updated on 03/Sep/20

	$Ok . Does that make the$ $question incorrect ?$
	Commented by Rasheed.Sindhi last updated on 03/Sep/20

	$O f c o u r s e!$
	Commented by pete last updated on 03/Sep/20

	$I appreciate your time sir .$
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