Question

in the arithmetic progression the fourth and sixth terms are 8 and 14 respectively. (i) first term (ii) common difference (j​

Answer 1

Step-by-step explanation:

Let " a " be the first term, " d " be the comman difference and " n " be the number of terms.

$\boxed{ \leadsto\sf a_{n} = a + (n - 1)d }$

$\sf The \: fourth \: term = 8$

$\sf \dashrightarrow a_{4} = 8$

$\sf \dashrightarrow a + 3d = 8.....(i)$

$\sf The \: sixth \: term = 14$

$\sf \dashrightarrow a_{6}= 14$

$\sf \dashrightarrow a + 5d = 14.....(ii)$

$\sf Subtracting \: equation \: (i) \: from \: (ii)$

$\sf \dashrightarrow a + 5d -(a + 3d) = 14 -8$

$\sf \dashrightarrow a + 5d - a - 3d = 6$

$\sf \dashrightarrow 2d = 6$

$\sf \dashrightarrow d = 3$

$\sf Substituting \: d = 3\: in \: equation \: (i)$

$\sf \dashrightarrow a + 3d = 8$

$\sf \dashrightarrow a + 3(3) = 8$

$\sf \dashrightarrow a + 9 = 8$

$\sf \dashrightarrow a = 8 -9 = -1$

$\sf \therefore\underline{The\: first \: term = -1}$

$\sf\therefore \underline{Common \: difference = 3}$

$\sf Now, \: the \: sum \: of \: 20\: terms \: of \: the \:AP:-$

$\boxed{ \leadsto\sf S_{n} = \frac{n}{2} \bigg[ 2 a + (n - 1)d \bigg] }$

$\dashrightarrow \sf S_{20} = \dfrac{20}{2} \bigg[ 2(-1) + (20 - 1)3\bigg ]$

$\dashrightarrow \sf S_{20} = 10(-2 + 19 \times 3)$

$\dashrightarrow \sf S_{20} = 10(-2 + 57)$

$\dashrightarrow \sf S_{20} = 10\times 55$

$\dashrightarrow \sf S_{20} = 550$

$\sf \therefore\underline{Sum \: of\: 20 \: terms = 550}$