Answer :

Let Monday be the first term i.e. a = t_{1}

Let Tuesday be the second term i.e t_{2}

Let Wednesday be the third term i.e t_{3}

Let Thursday be the fourth term i.e t_{4}

Let Friday be the fifth term i.e t_{5}

Let Saturday be the sixth term i.e t_{6}

Given: t_{1} + t_{6} = 5 + (t_{2} + t_{6})

⇒ a = 5 + (t_{2} + t_{6}) – t_{6}

⇒ a = 5 + t_{2} …..(1)

We know that,

Now, By using n^{th} term of an A.P. formula

t_{n} = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

Thus, t_{2} = a + (2 – 1) d

⇒ t_{2} = a + d

Now substitute value of t_{2} in (1) we get,

⇒ a = 5 + (a + d)

⇒ d = a – 5 – a = – 5

Given: t_{3} = – 30°

Thus, t_{3} = a + (3 – 1) × ( – 5)

⇒ – 30 = a + 2 × ( – 5)

⇒ – 30 = a – 10

⇒ a = – 30 + 10 = – 20°

Thus, Monday, a = t_{1} = – 20°

Using formula t_{n + 1} = t_{n} + d

We can find the value of the other terms

Tuesday, t_{2} = t_{1} + d = – 20 – 5 = – 25°

Wednesday, t_{3} = t_{2} + d = – 25 – 5 = – 30°

Thursday, t_{4} = t_{3} + d = – 30 – 5 = – 35°

Friday, t_{5} = t_{4} + d = – 35 – 5 = 40°

Saturday, t_{6} = t_{5} + d = – 40 – 5 = – 45°

Thus, we obtain an A.P.

– 20°, – 25°, – 30°, – 35°, – 40°, – 45°

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