Question

a^2 ×a^3×b^3×b^4/a^5×b^2​

Answer 1

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ANSWER

The given series are (2a−b),(4a−3b),(6a−5b),...n terms

∴a=(2a−b)

and d=((4a−3b)−(2a−b))=2a−2b=2(a−b)

∴S

n

=2a+(n−1)d

⟹S

n

=2(2a−b)+[(n−1)×2(a−b)]

⟹S

n

=(4a−2b)+[(2a−2b)×(n−1)]

=4a−2b−2a+2b+2an−2bn

=2a+2an−2bn

S

n

=2[a(2n+1)−bn]

Answer 2

Answer:

a² × a³× b³ × b⁴/a⁵ × b²

a(²+³) × b(³+⁴) /a⁵ × b². (when bases are same and in multiplic

a⁵ × b⁷/a⁵ × b². cation powers are added)

b⁷/b²

b⁵

therefore b⁵ is the ans