Question

In an arithmetic progression, if the 15th term is 18 and the 17th term is 14, then the sum of the first 50 terms is equal to
1)the 50th term
2)three times the 49th term.
3)five times the 48th term.
4)zero​

Answer 1

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Step-by-step explanation:

2)three times the 49th term..

Answer 2

$Let \: \pink {a} \:and \: \blue {d} \:are \\first \: term \: and \: Common \: differnce \: of \: an \:A.P$

$\boxed { \blue {n^{th} \:term (a_{n}) = a + (n-1)d }}$

$15^{th} \:term = 18 \:(given)$

$\implies a + 14d = 18 \: --(1)$

$17^{th} \:term = 14 \:(given)$

$\implies a + 16d = 14 \: --(2)$

/* Subtract equation (1) from equation (2), we get */

$\implies a + 16d - ( a + 14d) = 14 - 18$

$\implies a + 16d - a - 14d = -4$

$\implies 2d = -4$

$\implies d = \frac{-4}{2}$

$\implies d = -2$

/* Put d = -2 in equation (1) , we get */

$\implies a + 14(-2) = 18$

$\implies a - 28 = 18$

$\implies a = 18 + 28$

$\implies a = 46$

$\boxed { \pink { Sum \:of \:n \:terms (S_{n}) = \frac{n}{2}[2a + (n-1)d ] }}$

$Here, a = 46, d = -2 \: and \: n = 50$

$S_{50} = \frac{50}{2} [ 2\times 46 + (50-1)(-2)] \\= 25[ 92 + 49 \times (-2) ] \\= 25 ( 92 - 98 ) \\= 25 \times (-6) \\= - 150\: --(3)$

$i) 50^{th} \:term \: of \: A.P \\= a + 49d \\= 46 + 49 (-2) \\= 46 - 98 \\= 52\: --(4)$

$3 \:times \: 49^{th}\:term \\=3( a + 48d ) \\= 3[ 46 + 48(-2)] \\= 3[46 - 96 ] \\= 3 (-50) \\= -150\: --(5)$

Therefore.,

$\red{ S_{50} } \green {= 3 \:times \: 49^{th}\:term}$

$Option \: \orange {(2) } \:is \: correct.$

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