X1, x2... are in ap if x1+x7+x10=-6 and x3+x8+x12=-11 then x3+x8+x22=?

1.	X1, x2... are in ap if x1+x7+x10=-6 and x3+x8+x12=-11 then x3+x8+x22=?
Answer» $Given \: x_{1}, x_{2}, \cdot \cdot \cdot , are \:in \: A.P$ $Let \: a \: and \: d \: are \: first \: term \: and \\common \:difference \:of \: A.P$ $x_{1} = a , \: and \: d = x_{2} - x_{1}$ $\boxed { \pink { n^{th} \: term (a_{n}) = a+(n-1)d }}$ $i) x_{1} + x_{7} + x_{10} = -6$ $\implies a + (a+6d) + a+9d = -6$ $\implies 3a + 15d = -6\: ---(1)$ $ii) x_{3} + x_{8} + x_{12} = -11$ $\implies (a + 2d) + (a+7d) + (a+11d) = -11$ $\implies 3a + 20d = -11\: ---(2)$ /* SUBTRACT Equation (1) from equation (2) , we get / $\implies 5d = -<klux>5</klux>$ / DIVIDING both sides of equation by 5, we get / $\implies d = -1 \: --(3)$ / Put d = -1 in equation (1) , we get */ $3a + 15\times (-1) = -6$ $\implies 3a = -6 + 15$ $\implies 3a = 9$ $\implies a = \frac{9}{3} = 3\: ---(4)$ $Now, Value \:of \: x_{3} + x_{8} + x_{22} \\= (a+2d) + (a+7d) + (a+21d) \\= 3a + 30d \\= 3\times 3+ 30 \times (-1) \: [ From \:(3) \:and \:(4) ]$ $= 9 - 30 \\= -21$ Therefore., $\red {Value \:of \: x_{3} + x_{8} + x_{22}} \green {= -21 }$ •••♪ •••♪

Answer»

$Given \: x_{1}, x_{2}, \cdot \cdot \cdot , are \:in \: A.P$