Question

The 4th and 7th term of an arithmetic progression are 11 and -4 respectively. What is the 15th term?

A) -49 B) -44 C) -39 D) -34

Answer 1

$\large\bf\underline{Given:-}$

• 4th term = 11
• 7th term = -4

$\large\bf\underline {To \: find:-}$

• 15th term

$\huge\bf\underline{Solution:-}$

$\ddag \: \bf \: a_n = a + (n - 1)d$

• 4th term = a + 3d
• 7th term = a + 6d

$\rm \leadsto \: a + 3d = 11......(i) \\ \rm \leadsto \: a + 6d = - 4.......(ii)$

Solving eq.(i) and (ii), we get,

$\rm \: a + 3d = 11 \\ \rm \: a + 6d = - 4 \\ \rm \: - \: \: \: - \: \: \: \: \: \: \: \: \: + \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: \: \: \: \: \: \: - 3d = 15\\ \rm \: \: \: \: \: \: \: \: \: \: \: d = \frac{ - 15}{3} \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: d = - 5$

Substituting value of d = -5 in eq.(i)

$\rightarrowtail \rm \: a + 3d = 11 \\ \\ \rightarrowtail \rm \: a + 3 \times (- 5) = 11 \\ \\ \rightarrowtail \rm \: a - 15 = 11 \\ \\ \rightarrowtail \rm \: a = 11 + 15 \\ \\ \rightarrowtail \rm \: a = 26$

Now , finding 15th term :-

• ≫ 15th term = a + 14d

»» 15th term = 26 + 14 × (-5)

»» 15th term = 26 -70

»» 15th term = -44

Hence 15th term is -44.