Question

4th terms are in AP the sum of 2nd term and 3rd term is 22 and the product of first and 4th term 85 find the 4th term AP

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{a4=17}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

□a2+a3=22

□a2+a3=22□a1×a4=85

□a4=?

$=)a2+a3=22\\=)a+d+a+2d=22\\=)2a+3d=22\\=)3d=22-2a -----(1)\\again,\\=)a\times(a4)=85\\=)a\times(a+3d)=85\\=)a\times(a+22-2a)=85\\=)a\times(22-a)=85\\=)22a-a^{2}=85\\=)a^{2}-22a+85=0$

${a}^{2} - 17a - 5a + 85 = 0 \\ a(a - 17) - 5(a - 17) \\ (a - 5)(a - 17) = 0 \\ a = 5 \: and \: 17$

□a=5 and 17

>>at first we take a=5

>>putting value of a in (1)

$= )3d = 22 - 2a \\ = ) 3d = 22 - 2 \times 5 \\ = )3d = 22 - 10 \\ = )d = \frac{12}{3} \\ = )d = 4$

□d=4

$so \: a4 = a + 3d \\ = )5 + 3 \times 4 \\ = )5 + 12 \\ = )17$

>>a4=17 when we take a=5 and d=4

we take,

□a=17

>>putting value of a in (1)

$= )3d = 22 - 2a \\ = )3d = 22 - 2 \times 17 \\ = )3d = 22 - 34 \\ = )3d = - 14 \\ = )d = \frac{ - 14}{3}$

□d=-14/3

$so \:a4 = a + 3d \\ = )17 +( \frac{ - 14}{3} ) \\ = )17 - \frac{14}{ 3} \\ = ) \frac{51 - 14}{3} \\ = ) \frac{37}{ 3}$

□a4=37/3 when we take a=17 and d=-14/3.

>>so from this our conclusion is that we choose normal value of a4 and d.

$\huge{\boxed{\boxed{\sf{a4=17}}}}$