T 1 - t 3 -22 and t 1 - t 3 -85 find S10-

t_1+t_3=a+a+2d=2a+2d=22 a+d=11 t_1× t_3=a(a+2d)=(a+d d)(a+d+d)=(a+d)^2 (d)^2=11^2 d^2=85 d=± 6 d=6,a=5;d= 6,a=17 S_10=10/2[2(5)+(10 1)6] or ...

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Byju's Answer

Standard X

Mathematics

Sum of N Terms of an AP

t 1 + t 3 =22...

Question

$t_{1} + t_{3} = 22$ and $t_{1} \times t_{3} = 85$ find $S_{10}$ .

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Solution

$t_{1} + t_{3} = a + a + 2 d = 2 a + 2 d = 22 ⟹ a + d = 11$
$t_{1} \times t_{3} = a (a + 2 d) = (a + d - d) (a + d + d) = (a + d)^{2} - (d)^{2} = 11^{2} - d^{2} = 85 ⟹ d = \pm 6$
$d = 6, a = 5; d = - 6, a = 17$
$S_{10} = \frac{10}{2} [2 (5) + (10 - 1) 6] o r = \frac{10}{2} [2 (17) + (10 - 1) (- 6)]$
$S_{10} = 320 o r - 20$

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t1+t3=22 and t1×t3=85 find S10.

t1+t3=a+a+2d=2a+2d=22⟹a+d=11t1×t3=a(a+2d)=(a+d−d)(a+d+d)=(a+d)2−(d)2=112−d2=85⟹d=±6d=6,a=5;d=−6,a=17S10=102[2(5)+(10−1)6] or =102[2(17)+(10−1)(−6)]S10=320 or −20

$t_{1} + t_{3} = 22$ and $t_{1} \times t_{3} = 85$ find $S_{10}$ .