Question

How many terms are there in the A.P. whose first and fifth terms are - 14 and 2
respectively and the sum of the terms is 40 ?​

Answer 1

$a=-14\\a_5=2\\a_5=a+4d\\2=-14+4d\\4d=16\\d=4\\\\S_n=\frac{n}{2} [2a+(n-1)d]\\\\40=\frac{n}{2}[-28+(n-1)4]\\ \\40=n[-14+(n-1)2]\\40=n[-14+2n-2]\\40=n[-16+2n]\\40=2n^2-16n\\2n^2-16n-40=0\\n^2-8n-20=0\\n^2-10n+2n-20=0\\n(n-10)+2(n-10)=0\\(n-10)(n+2)=0\\n=10\\or,\\n=-2$

n = -2 is not possible as number of terms cannot be negative.

Hence, n = 10.

Answer 2

$\large\underline\mathrm\red{Given:-}$

• $\large\mathrm{First \: term \: = \: - 14}$

• $\large\mathrm{Fifth \: term \: = \: 2}$

• $\large\mathrm{Sum \: of \: terms \: = \: 40}$

$\large\underline\mathrm\red{To \: find}$

• $\large\mathrm{Number \: of \: terms \: ?}$

$\large\underline\mathrm\red{Solution}$

$\large\mathrm{⟹ First \: term \: (a) \: = \: - 14}$

$\large\mathrm{⟹ Fifth \: term \: (a + 4d) = 2}$

$\large\underline\mathrm\red{Substitution \: value \: of \: a}$

$\large\mathrm{⟹ - 14 + 4d = 2}$

$\large\mathrm{⟹ 4d = 2 + 14}$

$\large\mathrm{⟹ 4d = 16}$

$\large\mathrm{⟹ d = 4}$

$\large\underline\mathrm\red{So \: we \: get, \: common \: difference \: (d) \: = \: 4}$

$\large\mathrm{⟹ Sn = n/2 {2a + (n - 1)d}}$

$\large\mathrm{⟹ 40 = n/2 { - 14 × 2 + (n - 1)4 }}$

$\large\mathrm{⟹ 40 = n/2 {-28 + 4n - 4}}$

$\large\mathrm{⟹ 40 × 2 = n (- 32 + 4n)}$

$\large\mathrm{⟹ 80 = - 32n \: + \: {4n}^{2}}$

$\large\underline\mathrm\red{On \: divide \: both \: side \: by \: 4 \: we \: get,}$

$\large\mathrm{⟹ 20 = - 8n + {n}^{2}}$

$\large\mathrm{⟹ {n}^{2} - 8n - 20 = 0}$

$\large\mathrm{⟹ n + 2n - 10n - 20}$

$\large\mathrm{⟹ n (n + 2) - 10 (n + 2)}$

$\large\mathrm{⟹ (n - 10) (n + 2)}$

$\large\mathrm{⟹ n = 10 \:\:\: or \:\:\: n = - 2}$

• $\large\underline\mathrm\red{Number \: of \: terms \: cannot \: be \: negative.}$

$\large\mathrm{⟹ Number \: of \: term \: = \: 10}$

$\large\underline\mathrm\red{≫≫ So, \: there \: are \: 10 \: term \: in \: Ap.}$

__________________________________