Question

solve the solution 1+4+7+10.............+x = 590

Answer 1

See the attach file for ur solution

Answer 2

Answer:

x = 58

Explanation:

we have,

1+4+7+10+....+x = 590

1,4,7,10, ...is an A.P

first term (a) = 1

common difference (d) = a2-a1

= 4-1

d = 3

i ) Let sum of 'n' terms= Sn

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

$590=\frac{n}{2}[2\times1+(n-1)3]$

$\imples 1180 = n(2+3n-3)$

=> 1180=n(3n-1)

=> 1180=3n²-n

=> 3n²-n-1180=0

Splitting the middle term, we get

=> 3n²-60n+59n-1180=0

=> 3n(n-20)+59(n-20)=0

=> (n-20)(3n+59)=0

=> n-20 = 0 or 3n+59 = 0

=> n = 20 or n= -59/3

n should not be negative.

Therefore,

n = 20

ii) But , we know that

x = a+(n-1)d

= 1+(20-1)3

= 1+19×3

= 1 + 57

= 58

Therefore,

x = 58

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