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Question Number 73247 by Lontum Hans last updated on 09/Nov/19

	Answered by Kunal12588 last updated on 09/Nov/19
	$a_8 −a_4 =40 ....(1) a_8 =(3/2)a_4 .....(2) put a_8 in eq^n (1) (3/2)a_4 −a_4 =40 ⇒(a_4 /2)=40 ⇒a_4 =80 put a_4 in eq^n (2) a_8 =(3/2)×80 ⇒a_8 =120 a_4 =80 ⇒a_1 +3d=80 ....(3) a_8 =120 ⇒a_1 +7d=120 ....(4) eq^n (4)−eq^n (3) 4d=40 ⇒d=10 put d in eq^n (3) a_1 +30=80 ⇒a_1 =50 ans(1) first term, i.e. a_1 =50 ans(2) common difference, i.e. d=10 ans(3) a_9 =a_1 +8d=50+80=130 a_(10) =a_9 +d=130+10=140 a_(21) =a_1 +20d=50+200=250 a_(32) =a_1 +31d=50+310=360 a_(41) =a_1 +40d=50+400=450 ans(4) S_n =(n/2)×{2a+(n−1)d} S_4 =(4/2)×{2×50+3×10}=2(100+30)=260 S_5 =(5/2)×{2×50+4×10}=(5/2)(100+40)=350$
	$a_{8} - a_{4} = 40 . . . . (1)$ $a_{8} = \frac{3}{2} a_{4} . . . . . (2)$ $p u t a_{8} i n {e q}^{n} (1)$ $\frac{3}{2} a_{4} - a_{4} = 40$ $\Rightarrow \frac{a_{4}}{2} = 40$ $\Rightarrow a_{4} = 80$ $p u t a_{4} i n {e q}^{n} (2)$ $a_{8} = \frac{3}{2} \times 80$ $\Rightarrow a_{8} = 120$ $a_{4} = 80$ $\Rightarrow a_{1} + 3 d = 80 . . . . (3)$ $a_{8} = 120$ $\Rightarrow a_{1} + 7 d = 120 . . . . (4)$ ${e q}^{n} (4) - {e q}^{n} (3)$ $4 d = 40$ $\Rightarrow d = 10$ $p u t d i n {e q}^{n} (3)$ $a_{1} + 30 = 80$ $\Rightarrow a_{1} = 50$ $a n s (1) f i r s t t e r m, i . e . a_{1} = 50$ $a n s (2) c o m m o n d i f f e r e n c e, i . e . d = 10$ $a n s (3)$ $a_{9} = a_{1} + 8 d = 50 + 80 = 130$ $a_{10} = a_{9} + d = 130 + 10 = 140$ $a_{21} = a_{1} + 20 d = 50 + 200 = 250$ $a_{32} = a_{1} + 31 d = 50 + 310 = 360$ $a_{41} = a_{1} + 40 d = 50 + 400 = 450$ $a n s (4)$ $S_{n} = \frac{n}{2} \times {2 a + (n - 1) d}$ $S_{4} = \frac{4}{2} \times {2 \times 50 + 3 \times 10} = 2 (100 + 30) = 260$ $S_{5} = \frac{5}{2} \times {2 \times 50 + 4 \times 10} = \frac{5}{2} (100 + 40) = 350$
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