Question

An arithmetic progression 5,12,19,… has 50 terms. Find its last term. Hence find the sum of its last 15 terms.

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here

First term (a) = 5

Also,

Common difference,

d = 12 - 5 = 7

Also,

n = 50

Using Formula :-

a(n) = a + (n - 1)d

a(50) = 5 + (50 - 1) × 7

= 5 + 49 × 7

= 348

Here,

First term of last 15 terms be a(36),

a(36) = 5 + (36 - 1) × 7

= 5 + 35 × 7

= 5 + 245

= 250

So,

Sum of last 15 terms,

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

S(n) = 15/2{a × 250 + (15 - 1)7}

S(n) = 15/2{500 + 14 × 7}

S(n) = 15/2 × 598

S(n) = 4485

Therefore,

Sum of last 15 terms = 4485