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Sum = 3250
n2[40+(n−1)×15]=3250n2[40+(n−1)×15]=3250
on simplifying we get,
15n2+25n−6500⟹3n2+5n−1300⟹(n−20)(3n+65)∴n=0=0=0=20
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n2[40+(n−1)×15]=3250n2[40+(n−1)×15]=3250
on simplifying we get,
15n2+25n−6500⟹3n2+5n−1300⟹(n−20)(3n+65)∴n=0=0=0=20
I hope it find u help
Plz follow me
Answered by
2
Answer:
20 months
Step-by-step explanation:
this question forms an AP
where, a = 20
d = 15
and sn = 3250
sn = n/2 (2a + (n-1)d)
⇒3250 = n/2 (2×20 + (n-1) 15)
⇒15n² + 25n - 6500= 0
⇒n= 20 or -65/3
-65/3 is not possible as time can not be negative
therefore answer is 20 months
poojashettigar:
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