#### The 27^{th} term of the AP 298, 293, 288.is ____.

Given a_{1} = 298 and a_{2} = 293. ⇒ d = 293 – 298 = -5 we know that, a_{n} = a_{1} + (n – 1)d a_{27} = 298 + (27 – 1)(-5) = 298 +(26)(-5) = 298 – 130 ∴ a_{27} = 168

#### The common difference of AP −5, 0, 5, 10, 15, is ____.

Given, a_{1}=−5 and a_{2}=0 Common difference d= a_{2}−a_{1} =0− (−5) = 5 Hence, the common difference is 5.

#### A sequence is formed by selecting the even numbers in ascending order from the AP 1, 2, 3, 4, 5, 6, 7, 8,…….. The new sequence formed is also an AP.

Given AP is 1, 2, 3, 4, 5, 6, 7, 8… New sequence is 2, 4, 6, 8… 4 – 2 = 6 – 4 = 8 – 6 = ……. The difference between any two consecutive terms in the new sequence is a constant. Hence, the new sequence is also an AP.

#### If each term of the AP 1, 2, 3, is increased by 3, then the resulting sequence is also an AP.

Given, AP 1, 2, 3… New sequence formed by increasing each term of the given AP by 3 is 4, 5, 6 The difference between any two consecutive terms in the new sequence is constant Hence; the new sequence formed is an AP.

#### An AP contains 11 terms. If a_{1} = 2 and d = 3, then the value of its middle term is ____.

We know that, a_{n} = a_{1} + (n – 1)d ⇒ a_{6} = a_{1 } + (6 – 1)d = 2 + (6 -1)3 = 2 + 15 = 17 Hence, the value of the middle term of the given AP is 17.

#### The cost of a television is Rs. 20,000. If its value depreciates at Rs. 800 per year, then its value after 10 years is ____.

The prices of television at the starting of each year, forms an AP i.e., 20000, 19200, 18400… ⇒ a_{1} = 20000 and d = −800 We know that, ⇒ a_{n} = a_{1} + (n – 1)d ⇒ a _{11}= 20000 + (11 – 1)(-800) ⇒ a _{11}= 20000 − 8000 = 12000 Hence, the price of television after 10 years is Rs. 12,000.

#### If an AP, (a_{11})^{2} – (a_{1})^{2} = 1500 and (a_{11} – a_{1}) = 30, then a_{1} + a_{11} = ____.

Given, (a_{11})^{2} – (a_{1})^{2} = 1500 and (a_{11} – a_{1})= 30 ⇒ (a_{11} +a_{1})(a_{11} – a_{1}) = 1500 [∵ a^{2} –b^{2} = (a + b)(a – b)] ⇒ (a_{11} +a_{1})30 = 1500 ∴ a_{11} +a_{1} = 50

#### The sequence 13, 10, 7, 4… is an AP.

Given, a_{1} = 13, a_{2} = 10 and a_{3} = 7 a_{2} – a_{1} = 10 – 13 = -3 a_{3} – a_{2 } = 7 – 10 = -3 ⇒ a_{2} – a_{1} = a_{3} – a_{2} In an AP , the difference between any two consecutive terms is constant. Hence, the given sequence is an AP.

#### If each term of the AP 5, 8, 11, is increased by −2, then the resulting sequence is also an AP.

Given, AP is 5, 8, 11… New sequence formed by increasing each term of the given AP by −2 is 3, 6, 9, 12,.... ⇒ 6 − 3 = 9 − 6 = 12 − 9 ... The difference between any two consecutive terms in the new sequence is constant Hence, the new sequence formed is an AP.