Question

Hiw manybterms of series 1,4,7 are needed to sum 715

Answer 1

Hi ,

1 , 4 , 7 , ... is given series .

a2 - a1 = 4 -1 = 3

a3 - a2 = 7 - 4 = 3

Therefore , given series is an A.P

d = 3

Let the sum n terms = 715

n/2[ 2a + ( n - 1 )d ] = 715

n[ 2× 1 + ( n - 1 )3 ] = 715 × 2

n[ 2 + 3n - 3 ] = 1430

3n² - n - 1430 = 0

Compare above equation with ax²+bx+c = 0

we get ,

a = 3 , b = -1 , c = -1430

Discreaminant ( D ) = b² - 4ac

D = ( -1 )² - 4 × 3 × ( -1430 )

D = 1 + 17160

D = 17161

n = [ ( - b )± √D ]/2a

n = [ 1 ± √17161 ]/( 2 × 3 )

n = [ 1 ± 131 ]/ 6

n = 132/6 or n = -130/6

n = 22

Therefore ,

Sum of 22 terms = 715

I hope this helps you.

: )

Answer 2

Heya !!



AP = 1 , 4 , 7



First term ( A ) = 1



Common difference ( D ) = 4-1 = 3


Sn = 715



N/2 × [ 2A + ( N - 1 ) × D ] = 715


N/2 × [ 2 × 1 + ( N - 1 ) × 3 ] = 715


N/2 × ( 2 + 3N - 3 ) = 175



N/2 × ( -1 + 3N ) = 175




N ( -1 + 3N ) = 715 × 2


-N + 3N² = 1430



3N² - N - 1430 = 0




Discriminant ( D ) = B²-4AC



=> (-1)² - 4 × 3 × -1430




=> 1 + 17160


=> 17161




Root's= -B+ ✓D/2A or -B - ✓D/2A




=> -(-1) + ✓17161 / 2 × 3 or -(-1) - ✓17161/2×3




=> 1 + 131 / 6 or 1-131/6




=> 132/6 or -130/6


=> 22



N = 22


Hence,


22 terms should be taken to get the sum 715.