Question

find the sum of 51 terms of an ap whose second and third terms are 14 and 48

Answer 1

HEYA MATE, HERE IS UR ANSWER ______________________

a2=a+d
14=a+d________(1)

a3=a+2d
48=a+2d_______(2)

Eliminating 1 and 2

a+d=14
a+2d=48
- - -

-d=-34

d=34

a+d=14

a+34=14
a=14-34
a=-20

n=51

Sn=n/2 {2a+(n-1)d}

S51=51/2 {2×-20+(51-1)34}

S51=51/2 {-40+50×34}

S51=51/2(-40+1770)

S51=51/2×1660

S51=42,330

_____☆☆HOPE THE ANSWER HELPS YOU. ☆☆____☆☆BE BRAINLY ☆☆

Answer 2

Second term = 1 4
= > a + d = 14
= > a = 14 - d -------: ( 1 )



Third term = 48
= > a + 2d = 48



From ( i ) , a = 14 - d



= > 14 - d + 2d = 48


= > d = 34





In ( i ),

a = 14 - 34
a = - 20




So,
51th term = a + 50 d
51th term = -20 + 50( 34 )
51th term = - 20 + 1700
51th term = 1680




$\textbf{As we know}, \bold{ S_{n} =\frac{n}{2}[ n + l ] }$

$S_{50}= \frac{50}{2}[ 14 + 1680 ] \\ \\ \\ S_{50}= \frac{50}{2}[ 1694] \\ \\ \\ S_{50} = 25 \times 1694 \\ \\ \\ S_{50}= 42350$




Hence, sum of its 51 terms is 42350 - 20 = 42330